2018-04-03 19:23:33 +02:00
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2007-06-12 09:07:21 -04:00
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2007-08-29 09:11:44 -04:00
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2007-03-21 11:12:56 -04:00
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2009-04-02 16:46:06 -04:00
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2007-03-21 11:12:56 -04:00
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2007-08-29 09:11:44 -04:00
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2007-03-29 11:56:46 -04:00
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2007-04-19 21:01:03 -04:00
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2007-06-08 15:33:54 -04:00
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2007-08-29 09:11:44 -04:00
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2007-12-17 20:14:04 -05:00
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2007-12-18 16:15:09 -05:00
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2008-03-24 15:02:04 -04:00
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2009-01-15 13:51:03 -08:00
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include cleanup: Update gfp.h and slab.h includes to prepare for breaking implicit slab.h inclusion from percpu.h
percpu.h is included by sched.h and module.h and thus ends up being
included when building most .c files. percpu.h includes slab.h which
in turn includes gfp.h making everything defined by the two files
universally available and complicating inclusion dependencies.
percpu.h -> slab.h dependency is about to be removed. Prepare for
this change by updating users of gfp and slab facilities include those
headers directly instead of assuming availability. As this conversion
needs to touch large number of source files, the following script is
used as the basis of conversion.
http://userweb.kernel.org/~tj/misc/slabh-sweep.py
The script does the followings.
* Scan files for gfp and slab usages and update includes such that
only the necessary includes are there. ie. if only gfp is used,
gfp.h, if slab is used, slab.h.
* When the script inserts a new include, it looks at the include
blocks and try to put the new include such that its order conforms
to its surrounding. It's put in the include block which contains
core kernel includes, in the same order that the rest are ordered -
alphabetical, Christmas tree, rev-Xmas-tree or at the end if there
doesn't seem to be any matching order.
* If the script can't find a place to put a new include (mostly
because the file doesn't have fitting include block), it prints out
an error message indicating which .h file needs to be added to the
file.
The conversion was done in the following steps.
1. The initial automatic conversion of all .c files updated slightly
over 4000 files, deleting around 700 includes and adding ~480 gfp.h
and ~3000 slab.h inclusions. The script emitted errors for ~400
files.
2. Each error was manually checked. Some didn't need the inclusion,
some needed manual addition while adding it to implementation .h or
embedding .c file was more appropriate for others. This step added
inclusions to around 150 files.
3. The script was run again and the output was compared to the edits
from #2 to make sure no file was left behind.
4. Several build tests were done and a couple of problems were fixed.
e.g. lib/decompress_*.c used malloc/free() wrappers around slab
APIs requiring slab.h to be added manually.
5. The script was run on all .h files but without automatically
editing them as sprinkling gfp.h and slab.h inclusions around .h
files could easily lead to inclusion dependency hell. Most gfp.h
inclusion directives were ignored as stuff from gfp.h was usually
wildly available and often used in preprocessor macros. Each
slab.h inclusion directive was examined and added manually as
necessary.
6. percpu.h was updated not to include slab.h.
7. Build test were done on the following configurations and failures
were fixed. CONFIG_GCOV_KERNEL was turned off for all tests (as my
distributed build env didn't work with gcov compiles) and a few
more options had to be turned off depending on archs to make things
build (like ipr on powerpc/64 which failed due to missing writeq).
* x86 and x86_64 UP and SMP allmodconfig and a custom test config.
* powerpc and powerpc64 SMP allmodconfig
* sparc and sparc64 SMP allmodconfig
* ia64 SMP allmodconfig
* s390 SMP allmodconfig
* alpha SMP allmodconfig
* um on x86_64 SMP allmodconfig
8. percpu.h modifications were reverted so that it could be applied as
a separate patch and serve as bisection point.
Given the fact that I had only a couple of failures from tests on step
6, I'm fairly confident about the coverage of this conversion patch.
If there is a breakage, it's likely to be something in one of the arch
headers which should be easily discoverable easily on most builds of
the specific arch.
Signed-off-by: Tejun Heo <tj@kernel.org>
Guess-its-ok-by: Christoph Lameter <cl@linux-foundation.org>
Cc: Ingo Molnar <mingo@redhat.com>
Cc: Lee Schermerhorn <Lee.Schermerhorn@hp.com>
2010-03-24 17:04:11 +09:00
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2011-11-30 10:45:38 -05:00
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btrfs: Remove custom crc32c init code
The custom crc32 init code was introduced in
14a958e678cd ("Btrfs: fix btrfs boot when compiled as built-in") to
enable using btrfs as a built-in. However, later as pointed out by
60efa5eb2e88 ("Btrfs: use late_initcall instead of module_init") this
wasn't enough and finally btrfs was switched to late_initcall which
comes after the generic crc32c implementation is initiliased. The
latter commit superseeded the former. Now that we don't have to
maintain our own code let's just remove it and switch to using the
generic implementation.
Despite touching a lot of files the patch is really simple. Here is the gist of
the changes:
1. Select LIBCRC32C rather than the low-level modules.
2. s/btrfs_crc32c/crc32c/g
3. replace hash.h with linux/crc32c.h
4. Move the btrfs namehash funcs to ctree.h and change the tree accordingly.
I've tested this with btrfs being both a module and a built-in and xfstest
doesn't complain.
Does seem to fix the longstanding problem of not automatically selectiong
the crc32c module when btrfs is used. Possibly there is a workaround in
dracut.
The modinfo confirms that now all the module dependencies are there:
before:
depends: zstd_compress,zstd_decompress,raid6_pq,xor,zlib_deflate
after:
depends: libcrc32c,zstd_compress,zstd_decompress,raid6_pq,xor,zlib_deflate
Signed-off-by: Nikolay Borisov <nborisov@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
[ add more info to changelog from mails ]
Signed-off-by: David Sterba <dsterba@suse.com>
2018-01-08 11:45:05 +02:00
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2013-01-29 06:04:50 +00:00
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2023-08-25 16:19:29 -04:00
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2023-11-22 12:17:44 -05:00
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2022-10-19 10:50:49 -04:00
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btrfs: implement delayed inode items operation
Changelog V5 -> V6:
- Fix oom when the memory load is high, by storing the delayed nodes into the
root's radix tree, and letting btrfs inodes go.
Changelog V4 -> V5:
- Fix the race on adding the delayed node to the inode, which is spotted by
Chris Mason.
- Merge Chris Mason's incremental patch into this patch.
- Fix deadlock between readdir() and memory fault, which is reported by
Itaru Kitayama.
Changelog V3 -> V4:
- Fix nested lock, which is reported by Itaru Kitayama, by updating space cache
inode in time.
Changelog V2 -> V3:
- Fix the race between the delayed worker and the task which does delayed items
balance, which is reported by Tsutomu Itoh.
- Modify the patch address David Sterba's comment.
- Fix the bug of the cpu recursion spinlock, reported by Chris Mason
Changelog V1 -> V2:
- break up the global rb-tree, use a list to manage the delayed nodes,
which is created for every directory and file, and used to manage the
delayed directory name index items and the delayed inode item.
- introduce a worker to deal with the delayed nodes.
Compare with Ext3/4, the performance of file creation and deletion on btrfs
is very poor. the reason is that btrfs must do a lot of b+ tree insertions,
such as inode item, directory name item, directory name index and so on.
If we can do some delayed b+ tree insertion or deletion, we can improve the
performance, so we made this patch which implemented delayed directory name
index insertion/deletion and delayed inode update.
Implementation:
- introduce a delayed root object into the filesystem, that use two lists to
manage the delayed nodes which are created for every file/directory.
One is used to manage all the delayed nodes that have delayed items. And the
other is used to manage the delayed nodes which is waiting to be dealt with
by the work thread.
- Every delayed node has two rb-tree, one is used to manage the directory name
index which is going to be inserted into b+ tree, and the other is used to
manage the directory name index which is going to be deleted from b+ tree.
- introduce a worker to deal with the delayed operation. This worker is used
to deal with the works of the delayed directory name index items insertion
and deletion and the delayed inode update.
When the delayed items is beyond the lower limit, we create works for some
delayed nodes and insert them into the work queue of the worker, and then
go back.
When the delayed items is beyond the upper bound, we create works for all
the delayed nodes that haven't been dealt with, and insert them into the work
queue of the worker, and then wait for that the untreated items is below some
threshold value.
- When we want to insert a directory name index into b+ tree, we just add the
information into the delayed inserting rb-tree.
And then we check the number of the delayed items and do delayed items
balance. (The balance policy is above.)
- When we want to delete a directory name index from the b+ tree, we search it
in the inserting rb-tree at first. If we look it up, just drop it. If not,
add the key of it into the delayed deleting rb-tree.
Similar to the delayed inserting rb-tree, we also check the number of the
delayed items and do delayed items balance.
(The same to inserting manipulation)
- When we want to update the metadata of some inode, we cached the data of the
inode into the delayed node. the worker will flush it into the b+ tree after
dealing with the delayed insertion and deletion.
- We will move the delayed node to the tail of the list after we access the
delayed node, By this way, we can cache more delayed items and merge more
inode updates.
- If we want to commit transaction, we will deal with all the delayed node.
- the delayed node will be freed when we free the btrfs inode.
- Before we log the inode items, we commit all the directory name index items
and the delayed inode update.
I did a quick test by the benchmark tool[1] and found we can improve the
performance of file creation by ~15%, and file deletion by ~20%.
Before applying this patch:
Create files:
Total files: 50000
Total time: 1.096108
Average time: 0.000022
Delete files:
Total files: 50000
Total time: 1.510403
Average time: 0.000030
After applying this patch:
Create files:
Total files: 50000
Total time: 0.932899
Average time: 0.000019
Delete files:
Total files: 50000
Total time: 1.215732
Average time: 0.000024
[1] http://marc.info/?l=linux-btrfs&m=128212635122920&q=p3
Many thanks for Kitayama-san's help!
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Reviewed-by: David Sterba <dave@jikos.cz>
Tested-by: Tsutomu Itoh <t-itoh@jp.fujitsu.com>
Tested-by: Itaru Kitayama <kitayama@cl.bb4u.ne.jp>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-04-22 18:12:22 +08:00
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2007-03-21 11:12:56 -04:00
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2007-03-22 12:13:20 -04:00
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2007-03-23 10:01:08 -04:00
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2007-04-02 10:50:19 -04:00
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2024-06-25 15:52:28 +01:00
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Btrfs: add support for inode properties
This change adds infrastructure to allow for generic properties for
inodes. Properties are name/value pairs that can be associated with
inodes for different purposes. They are stored as xattrs with the
prefix "btrfs."
Properties can be inherited - this means when a directory inode has
inheritable properties set, these are added to new inodes created
under that directory. Further, subvolumes can also have properties
associated with them, and they can be inherited from their parent
subvolume. Naturally, directory properties have priority over subvolume
properties (in practice a subvolume property is just a regular
property associated with the root inode, objectid 256, of the
subvolume's fs tree).
This change also adds one specific property implementation, named
"compression", whose values can be "lzo" or "zlib" and it's an
inheritable property.
The corresponding changes to btrfs-progs were also implemented.
A patch with xfstests for this feature will follow once there's
agreement on this change/feature.
Further, the script at the bottom of this commit message was used to
do some benchmarks to measure any performance penalties of this feature.
Basically the tests correspond to:
Test 1 - create a filesystem and mount it with compress-force=lzo,
then sequentially create N files of 64Kb each, measure how long it took
to create the files, unmount the filesystem, mount the filesystem and
perform an 'ls -lha' against the test directory holding the N files, and
report the time the command took.
Test 2 - create a filesystem and don't use any compression option when
mounting it - instead set the compression property of the subvolume's
root to 'lzo'. Then create N files of 64Kb, and report the time it took.
The unmount the filesystem, mount it again and perform an 'ls -lha' like
in the former test. This means every single file ends up with a property
(xattr) associated to it.
Test 3 - same as test 2, but uses 4 properties - 3 are duplicates of the
compression property, have no real effect other than adding more work
when inheriting properties and taking more btree leaf space.
Test 4 - same as test 3 but with 10 properties per file.
Results (in seconds, and averages of 5 runs each), for different N
numbers of files follow.
* Without properties (test 1)
file creation time ls -lha time
10 000 files 3.49 0.76
100 000 files 47.19 8.37
1 000 000 files 518.51 107.06
* With 1 property (compression property set to lzo - test 2)
file creation time ls -lha time
10 000 files 3.63 0.93
100 000 files 48.56 9.74
1 000 000 files 537.72 125.11
* With 4 properties (test 3)
file creation time ls -lha time
10 000 files 3.94 1.20
100 000 files 52.14 11.48
1 000 000 files 572.70 142.13
* With 10 properties (test 4)
file creation time ls -lha time
10 000 files 4.61 1.35
100 000 files 58.86 13.83
1 000 000 files 656.01 177.61
The increased latencies with properties are essencialy because of:
*) When creating an inode, we now synchronously write 1 more item
(an xattr item) for each property inherited from the parent dir
(or subvolume). This could be done in an asynchronous way such
as we do for dir intex items (delayed-inode.c), which could help
reduce the file creation latency;
*) With properties, we now have larger fs trees. For this particular
test each xattr item uses 75 bytes of leaf space in the fs tree.
This could be less by using a new item for xattr items, instead of
the current btrfs_dir_item, since we could cut the 'location' and
'type' fields (saving 18 bytes) and maybe 'transid' too (saving a
total of 26 bytes per xattr item) from the btrfs_dir_item type.
Also tried batching the xattr insertions (ignoring proper hash
collision handling, since it didn't exist) when creating files that
inherit properties from their parent inode/subvolume, but the end
results were (surprisingly) essentially the same.
Test script:
$ cat test.pl
#!/usr/bin/perl -w
use strict;
use Time::HiRes qw(time);
use constant NUM_FILES => 10_000;
use constant FILE_SIZES => (64 * 1024);
use constant DEV => '/dev/sdb4';
use constant MNT_POINT => '/home/fdmanana/btrfs-tests/dev';
use constant TEST_DIR => (MNT_POINT . '/testdir');
system("mkfs.btrfs", "-l", "16384", "-f", DEV) == 0 or die "mkfs.btrfs failed!";
# following line for testing without properties
#system("mount", "-o", "compress-force=lzo", DEV, MNT_POINT) == 0 or die "mount failed!";
# following 2 lines for testing with properties
system("mount", DEV, MNT_POINT) == 0 or die "mount failed!";
system("btrfs", "prop", "set", MNT_POINT, "compression", "lzo") == 0 or die "set prop failed!";
system("mkdir", TEST_DIR) == 0 or die "mkdir failed!";
my ($t1, $t2);
$t1 = time();
for (my $i = 1; $i <= NUM_FILES; $i++) {
my $p = TEST_DIR . '/file_' . $i;
open(my $f, '>', $p) or die "Error opening file!";
$f->autoflush(1);
for (my $j = 0; $j < FILE_SIZES; $j += 4096) {
print $f ('A' x 4096) or die "Error writing to file!";
}
close($f);
}
$t2 = time();
print "Time to create " . NUM_FILES . ": " . ($t2 - $t1) . " seconds.\n";
system("umount", DEV) == 0 or die "umount failed!";
system("mount", DEV, MNT_POINT) == 0 or die "mount failed!";
$t1 = time();
system("bash -c 'ls -lha " . TEST_DIR . " > /dev/null'") == 0 or die "ls failed!";
$t2 = time();
print "Time to ls -lha all files: " . ($t2 - $t1) . " seconds.\n";
system("umount", DEV) == 0 or die "umount failed!";
Signed-off-by: Filipe David Borba Manana <fdmanana@gmail.com>
Signed-off-by: Josef Bacik <jbacik@fb.com>
Signed-off-by: Chris Mason <clm@fb.com>
2014-01-07 11:47:46 +00:00
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2007-11-16 11:45:54 -05:00
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2022-11-15 10:44:05 +01:00
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2008-07-21 02:01:56 +05:30
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Btrfs: Add zlib compression support
This is a large change for adding compression on reading and writing,
both for inline and regular extents. It does some fairly large
surgery to the writeback paths.
Compression is off by default and enabled by mount -o compress. Even
when the -o compress mount option is not used, it is possible to read
compressed extents off the disk.
If compression for a given set of pages fails to make them smaller, the
file is flagged to avoid future compression attempts later.
* While finding delalloc extents, the pages are locked before being sent down
to the delalloc handler. This allows the delalloc handler to do complex things
such as cleaning the pages, marking them writeback and starting IO on their
behalf.
* Inline extents are inserted at delalloc time now. This allows us to compress
the data before inserting the inline extent, and it allows us to insert
an inline extent that spans multiple pages.
* All of the in-memory extent representations (extent_map.c, ordered-data.c etc)
are changed to record both an in-memory size and an on disk size, as well
as a flag for compression.
From a disk format point of view, the extent pointers in the file are changed
to record the on disk size of a given extent and some encoding flags.
Space in the disk format is allocated for compression encoding, as well
as encryption and a generic 'other' field. Neither the encryption or the
'other' field are currently used.
In order to limit the amount of data read for a single random read in the
file, the size of a compressed extent is limited to 128k. This is a
software only limit, the disk format supports u64 sized compressed extents.
In order to limit the ram consumed while processing extents, the uncompressed
size of a compressed extent is limited to 256k. This is a software only limit
and will be subject to tuning later.
Checksumming is still done on compressed extents, and it is done on the
uncompressed version of the data. This way additional encodings can be
layered on without having to figure out which encoding to checksum.
Compression happens at delalloc time, which is basically singled threaded because
it is usually done by a single pdflush thread. This makes it tricky to
spread the compression load across all the cpus on the box. We'll have to
look at parallel pdflush walks of dirty inodes at a later time.
Decompression is hooked into readpages and it does spread across CPUs nicely.
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2008-10-29 14:49:59 -04:00
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2012-11-06 13:15:27 +01:00
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2013-03-15 09:47:08 -04:00
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2013-08-09 13:25:36 +08:00
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2019-06-18 16:09:16 -04:00
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2019-08-01 17:34:41 +02:00
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2020-11-10 20:26:08 +09:00
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2013-08-14 15:05:12 -04:00
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2019-06-20 15:37:44 -04:00
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2019-12-13 16:22:14 -08:00
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2014-07-17 12:39:00 -07:00
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2022-06-01 17:46:59 +08:00
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2022-10-19 10:50:47 -04:00
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2022-10-19 10:51:00 -04:00
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2022-10-26 15:08:25 -04:00
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2022-10-26 15:08:26 -04:00
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2022-10-26 15:08:29 -04:00
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2022-10-26 15:08:35 -04:00
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2022-10-26 15:08:37 -04:00
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2022-10-26 15:08:39 -04:00
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2022-12-15 16:06:31 -08:00
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Btrfs: add initial tracepoint support for btrfs
Tracepoints can provide insight into why btrfs hits bugs and be greatly
helpful for debugging, e.g
dd-7822 [000] 2121.641088: btrfs_inode_request: root = 5(FS_TREE), gen = 4, ino = 256, blocks = 8, disk_i_size = 0, last_trans = 8, logged_trans = 0
dd-7822 [000] 2121.641100: btrfs_inode_new: root = 5(FS_TREE), gen = 8, ino = 257, blocks = 0, disk_i_size = 0, last_trans = 0, logged_trans = 0
btrfs-transacti-7804 [001] 2146.935420: btrfs_cow_block: root = 2(EXTENT_TREE), refs = 2, orig_buf = 29368320 (orig_level = 0), cow_buf = 29388800 (cow_level = 0)
btrfs-transacti-7804 [001] 2146.935473: btrfs_cow_block: root = 1(ROOT_TREE), refs = 2, orig_buf = 29364224 (orig_level = 0), cow_buf = 29392896 (cow_level = 0)
btrfs-transacti-7804 [001] 2146.972221: btrfs_transaction_commit: root = 1(ROOT_TREE), gen = 8
flush-btrfs-2-7821 [001] 2155.824210: btrfs_chunk_alloc: root = 3(CHUNK_TREE), offset = 1103101952, size = 1073741824, num_stripes = 1, sub_stripes = 0, type = DATA
flush-btrfs-2-7821 [001] 2155.824241: btrfs_cow_block: root = 2(EXTENT_TREE), refs = 2, orig_buf = 29388800 (orig_level = 0), cow_buf = 29396992 (cow_level = 0)
flush-btrfs-2-7821 [001] 2155.824255: btrfs_cow_block: root = 4(DEV_TREE), refs = 2, orig_buf = 29372416 (orig_level = 0), cow_buf = 29401088 (cow_level = 0)
flush-btrfs-2-7821 [000] 2155.824329: btrfs_cow_block: root = 3(CHUNK_TREE), refs = 2, orig_buf = 20971520 (orig_level = 0), cow_buf = 20975616 (cow_level = 0)
btrfs-endio-wri-7800 [001] 2155.898019: btrfs_cow_block: root = 5(FS_TREE), refs = 2, orig_buf = 29384704 (orig_level = 0), cow_buf = 29405184 (cow_level = 0)
btrfs-endio-wri-7800 [001] 2155.898043: btrfs_cow_block: root = 7(CSUM_TREE), refs = 2, orig_buf = 29376512 (orig_level = 0), cow_buf = 29409280 (cow_level = 0)
Here is what I have added:
1) ordere_extent:
btrfs_ordered_extent_add
btrfs_ordered_extent_remove
btrfs_ordered_extent_start
btrfs_ordered_extent_put
These provide critical information to understand how ordered_extents are
updated.
2) extent_map:
btrfs_get_extent
extent_map is used in both read and write cases, and it is useful for tracking
how btrfs specific IO is running.
3) writepage:
__extent_writepage
btrfs_writepage_end_io_hook
Pages are cirtical resourses and produce a lot of corner cases during writeback,
so it is valuable to know how page is written to disk.
4) inode:
btrfs_inode_new
btrfs_inode_request
btrfs_inode_evict
These can show where and when a inode is created, when a inode is evicted.
5) sync:
btrfs_sync_file
btrfs_sync_fs
These show sync arguments.
6) transaction:
btrfs_transaction_commit
In transaction based filesystem, it will be useful to know the generation and
who does commit.
7) back reference and cow:
btrfs_delayed_tree_ref
btrfs_delayed_data_ref
btrfs_delayed_ref_head
btrfs_cow_block
Btrfs natively supports back references, these tracepoints are helpful on
understanding btrfs's COW mechanism.
8) chunk:
btrfs_chunk_alloc
btrfs_chunk_free
Chunk is a link between physical offset and logical offset, and stands for space
infomation in btrfs, and these are helpful on tracing space things.
9) reserved_extent:
btrfs_reserved_extent_alloc
btrfs_reserved_extent_free
These can show how btrfs uses its space.
Signed-off-by: Liu Bo <liubo2009@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-03-24 11:18:59 +00:00
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2009-09-21 17:01:09 -07:00
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2011-07-25 15:55:42 -04:00
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btrfs: allow mounting btrfs subvolumes with different ro/rw options
Given the following /etc/fstab entries:
/dev/sda3 /mnt/foo btrfs subvol=foo,ro 0 0
/dev/sda3 /mnt/bar btrfs subvol=bar,rw 0 0
you can't issue:
$ mount /mnt/foo
$ mount /mnt/bar
You would have to do:
$ mount /mnt/foo
$ mount -o remount,rw /mnt/foo
$ mount --bind -o remount,ro /mnt/foo
$ mount /mnt/bar
or
$ mount /mnt/bar
$ mount --rw /mnt/foo
$ mount --bind -o remount,ro /mnt/foo
With this patch you can do
$ mount /mnt/foo
$ mount /mnt/bar
$ cat /proc/self/mountinfo
49 33 0:41 /foo /mnt/foo ro,relatime shared:36 - btrfs /dev/sda3 rw,ssd,space_cache
87 33 0:41 /bar /mnt/bar rw,relatime shared:74 - btrfs /dev/sda3 rw,ssd,space_cache
Signed-off-by: Chris Mason <clm@fb.com>
2013-11-19 11:36:05 +01:00
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2009-01-05 21:25:51 -05:00
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2007-03-22 12:13:20 -04:00
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2023-11-02 07:54:50 +10:30
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2007-03-22 12:13:20 -04:00
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2007-03-21 11:12:56 -04:00
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2023-11-22 12:17:45 -05:00
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2024-07-13 18:33:44 +09:30
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2023-11-22 12:17:45 -05:00
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2025-01-30 18:58:19 +01:00
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2023-11-22 12:17:48 -05:00
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2023-11-22 12:17:45 -05:00
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2025-08-13 03:00:06 +09:00
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2007-08-29 09:11:44 -04:00
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2023-11-22 12:17:52 -05:00
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2018-03-09 14:37:01 +01:00
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2023-11-22 12:17:52 -05:00
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2018-03-09 14:37:01 +01:00
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2023-11-22 12:17:52 -05:00
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2019-12-13 16:22:14 -08:00
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2018-03-09 14:37:01 +01:00
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2023-11-22 12:17:52 -05:00
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2018-03-09 14:37:01 +01:00
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2023-11-22 12:17:52 -05:00
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2018-03-09 14:37:01 +01:00
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2018-05-21 17:07:19 -07:00
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2018-03-09 14:37:01 +01:00
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2023-11-22 12:17:52 -05:00
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2018-03-09 14:37:01 +01:00
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2024-05-21 19:27:31 +09:30
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2018-03-09 14:37:01 +01:00
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2020-06-04 15:18:06 +08:00
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2018-03-09 14:37:01 +01:00
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2023-11-22 12:17:52 -05:00
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2015-09-23 14:54:14 -04:00
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2023-11-22 12:17:44 -05:00
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2017-09-29 15:43:48 -04:00
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2025-08-12 16:04:41 -07:00
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2015-09-23 14:54:14 -04:00
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2012-01-16 22:04:48 +02:00
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2007-08-29 09:11:44 -04:00
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2023-11-22 12:17:44 -05:00
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2018-03-09 14:37:01 +01:00
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2023-11-22 12:17:44 -05:00
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2020-06-04 15:18:06 +08:00
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2023-11-22 12:17:44 -05:00
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2024-06-14 13:52:30 +09:30
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2024-06-14 13:52:31 +09:30
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2023-11-22 12:17:44 -05:00
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2024-06-14 13:52:30 +09:30
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2024-06-14 13:52:31 +09:30
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2023-11-22 12:17:44 -05:00
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2024-06-14 13:52:30 +09:30
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2024-06-14 13:52:31 +09:30
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2023-11-22 12:17:44 -05:00
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2018-03-09 14:37:01 +01:00
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2015-09-23 14:54:14 -04:00
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2023-11-22 12:17:44 -05:00
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2023-11-22 12:17:50 -05:00
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2023-11-22 12:17:44 -05:00
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2024-05-21 19:27:31 +09:30
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2023-11-22 12:17:44 -05:00
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2025-08-12 16:04:41 -07:00
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2023-11-22 12:17:44 -05:00
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2015-09-23 14:54:14 -04:00
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2023-11-22 12:17:44 -05:00
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2007-08-29 09:11:44 -04:00
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2025-06-02 17:53:19 +02:00
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2025-06-02 17:53:18 +02:00
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2025-06-02 17:53:19 +02:00
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2025-08-25 19:56:26 +09:30
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2025-06-02 17:53:19 +02:00
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2025-06-02 17:53:18 +02:00
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2025-06-02 17:53:19 +02:00
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2025-06-02 17:53:18 +02:00
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2025-08-25 19:56:26 +09:30
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2025-06-02 17:53:18 +02:00
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btrfs: accept and ignore compression level for lzo
The compression level is meaningless for lzo, but before commit
3f093ccb95f30 ("btrfs: harden parsing of compression mount options"),
it was silently ignored if passed.
After that commit, passing a level with lzo fails to mount:
BTRFS error: unrecognized compression value lzo:1
It seems reasonable for users to expect that lzo would permit a numeric
level option, as all the other algos do, even though the kernel's
implementation of LZO currently only supports a single level. Because it
has always worked to pass a level, it seems likely to me that users in
the real world are relying on doing so.
This patch restores the old behavior, giving "lzo:N" the same semantics
as all of the other compression algos.
To be clear, silly variants like "lzo:one", "lzo:the_first_option", or
"lzo:armageddon" also used to work. This isn't meant to suggest that
any possible mis-interpretation of mount options that once worked must
continue to work forever. This is an exceptional case where it makes
sense to preserve compatibility, both because the mis-interpretation is
reasonable, and because nothing tangible is sacrificed.
Finally update btrfs_show_options() to ignore the level of LZO, as it
is only the default level without any extra meaning.
Fixes: 3f093ccb95f30 ("btrfs: harden parsing of compression mount options")
Reviewed-by: Daniel Vacek <neelx@suse.com>
Reviewed-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: Calvin Owens <calvin@wbinvd.org>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2025-08-25 18:32:04 +09:30
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2025-06-02 17:53:18 +02:00
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2025-08-25 19:56:26 +09:30
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|
btrfs: accept and ignore compression level for lzo
The compression level is meaningless for lzo, but before commit
3f093ccb95f30 ("btrfs: harden parsing of compression mount options"),
it was silently ignored if passed.
After that commit, passing a level with lzo fails to mount:
BTRFS error: unrecognized compression value lzo:1
It seems reasonable for users to expect that lzo would permit a numeric
level option, as all the other algos do, even though the kernel's
implementation of LZO currently only supports a single level. Because it
has always worked to pass a level, it seems likely to me that users in
the real world are relying on doing so.
This patch restores the old behavior, giving "lzo:N" the same semantics
as all of the other compression algos.
To be clear, silly variants like "lzo:one", "lzo:the_first_option", or
"lzo:armageddon" also used to work. This isn't meant to suggest that
any possible mis-interpretation of mount options that once worked must
continue to work forever. This is an exceptional case where it makes
sense to preserve compatibility, both because the mis-interpretation is
reasonable, and because nothing tangible is sacrificed.
Finally update btrfs_show_options() to ignore the level of LZO, as it
is only the default level without any extra meaning.
Fixes: 3f093ccb95f30 ("btrfs: harden parsing of compression mount options")
Reviewed-by: Daniel Vacek <neelx@suse.com>
Reviewed-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: Calvin Owens <calvin@wbinvd.org>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2025-08-25 18:32:04 +09:30
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2025-06-02 17:53:18 +02:00
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2025-06-02 17:53:19 +02:00
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2025-06-02 17:53:18 +02:00
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2025-08-25 19:56:26 +09:30
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2025-06-02 17:53:18 +02:00
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2025-06-02 17:53:19 +02:00
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2025-06-02 17:53:18 +02:00
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2025-08-25 19:56:26 +09:30
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2025-06-02 17:53:18 +02:00
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2025-08-25 19:56:26 +09:30
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2025-06-02 17:53:18 +02:00
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2023-11-22 12:17:45 -05:00
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2025-06-11 12:02:59 +02:00
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2023-11-22 12:17:45 -05:00
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2025-06-02 17:53:18 +02:00
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2023-11-22 12:17:45 -05:00
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2024-05-21 19:27:31 +09:30
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2023-11-22 12:17:45 -05:00
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2023-11-22 12:17:54 -05:00
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2023-11-22 12:17:45 -05:00
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2025-05-09 19:26:31 +09:00
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2023-11-22 12:17:45 -05:00
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2024-06-14 13:52:30 +09:30
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2024-06-14 13:52:31 +09:30
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2023-11-22 12:17:45 -05:00
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2024-06-14 13:52:30 +09:30
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2024-06-14 13:52:31 +09:30
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2023-11-22 12:17:45 -05:00
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2025-08-12 16:04:41 -07:00
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2023-11-22 12:17:45 -05:00
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2023-11-22 12:17:53 -05:00
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2024-05-30 19:14:12 +02:00
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2024-07-13 18:33:44 +09:30
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2020-10-16 11:29:13 -04:00
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2023-11-22 12:17:47 -05:00
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2020-10-16 11:29:13 -04:00
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2024-07-13 18:33:44 +09:30
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2023-11-22 12:17:50 -05:00
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2020-06-04 15:18:06 +08:00
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2023-11-22 12:17:38 -05:00
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2020-06-04 15:18:06 +08:00
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2023-11-22 12:17:38 -05:00
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2023-11-22 12:17:47 -05:00
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2024-06-14 13:52:30 +09:30
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2024-06-14 13:52:31 +09:30
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2023-11-22 12:17:38 -05:00
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2020-06-04 15:18:06 +08:00
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2016-06-22 18:54:23 -04:00
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2023-11-22 12:17:47 -05:00
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2023-11-22 12:17:38 -05:00
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2023-04-27 09:45:32 +08:00
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2023-11-22 12:17:47 -05:00
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2023-11-22 12:17:38 -05:00
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2015-09-29 20:50:38 -07:00
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2023-11-22 12:17:38 -05:00
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2023-11-22 12:17:47 -05:00
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2023-11-22 12:17:38 -05:00
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2024-07-29 11:05:48 -04:00
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2022-08-23 17:28:20 +02:00
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2023-11-22 12:17:38 -05:00
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2009-11-07 06:19:16 +00:00
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2008-06-10 10:40:29 -04:00
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2023-11-22 12:17:40 -05:00
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2008-06-10 10:40:29 -04:00
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2023-11-22 12:17:40 -05:00
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2008-06-10 10:40:29 -04:00
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2023-11-22 12:17:40 -05:00
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2008-06-10 10:40:29 -04:00
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|
btrfs: force free space tree for bs > ps cases
[BUG]
Currently we only enforcing the free space tree for bs < ps cases, but
with the recently added bs > ps support, we lack the free space tree
enforcing, causing explicit v1 cache mount option to fail on bs > ps
cases:
# mount -o space_cache=v1 /dev/test/scratch1 /mnt/btrfs/
mount: /mnt/btrfs: wrong fs type, bad option, bad superblock on /dev/mapper/test-scratch1, missing codepage or helper program, or other error.
dmesg(1) may have more information after failed mount system call.
# dmesg -t | tail -n7
BTRFS: device fsid ac14a6fa-4ec9-449e-aec9-7d1777bfdc06 devid 1 transid 11 /dev/mapper/test-scratch1 (253:3) scanned by mount (2849)
BTRFS info (device dm-3): first mount of filesystem ac14a6fa-4ec9-449e-aec9-7d1777bfdc06
BTRFS info (device dm-3): using crc32c checksum algorithm
BTRFS warning (device dm-3): support for block size 8192 with page size 4096 is experimental, some features may be missing
BTRFS warning (device dm-3): space cache v1 is being deprecated and will be removed in a future release, please use -o space_cache=v2
BTRFS warning (device dm-3): v1 space cache is not supported for page size 4096 with sectorsize 8192
BTRFS error (device dm-3): open_ctree failed: -22
[FIX]
Just enable the same free space tree for bs > ps cases, aligning the
behavior to bs < ps cases.
Reviewed-by: Filipe Manana <fdmanana@suse.com>
Signed-off-by: Qu Wenruo <wqu@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2025-12-18 15:15:29 +10:30
|
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|
btrfs: only enforce free space tree if v1 cache is required for bs < ps cases
[BUG]
Since the introduction of btrfs bs < ps support, v1 cache was never on
the plan due to its hard coded PAGE_SIZE usage, and the future plan to
properly deprecate it.
However for bs < ps cases, even if 'nospace_cache,clear_cache' mount
option is specified, it's never respected and free space tree is always
enabled:
mkfs.btrfs -f -O ^bgt,fst $dev
mount $dev $mnt -o clear_cache,nospace_cache
umount $mnt
btrfs ins dump-super $dev
...
compat_ro_flags 0x3
( FREE_SPACE_TREE |
FREE_SPACE_TREE_VALID )
...
This means a different behavior compared to bs >= ps cases.
[CAUSE]
The forcing usage of v2 space cache is done inside
btrfs_set_free_space_cache_settings(), however it never checks if we're
even using space cache but always enabling v2 cache.
[FIX]
Instead unconditionally enable v2 cache, only forcing v2 cache if the
old v1 cache is required.
Now v2 space cache can be properly disabled on bs < ps cases:
mkfs.btrfs -f -O ^bgt,fst $dev
mount $dev $mnt -o clear_cache,nospace_cache
umount $mnt
btrfs ins dump-super $dev
...
compat_ro_flags 0x0
...
Fixes: 9f73f1aef98b ("btrfs: force v2 space cache usage for subpage mount")
Reviewed-by: Filipe Manana <fdmanana@suse.com>
Signed-off-by: Qu Wenruo <wqu@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2025-12-18 15:15:28 +10:30
|
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2023-11-22 12:17:40 -05:00
|
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|
btrfs: only enforce free space tree if v1 cache is required for bs < ps cases
[BUG]
Since the introduction of btrfs bs < ps support, v1 cache was never on
the plan due to its hard coded PAGE_SIZE usage, and the future plan to
properly deprecate it.
However for bs < ps cases, even if 'nospace_cache,clear_cache' mount
option is specified, it's never respected and free space tree is always
enabled:
mkfs.btrfs -f -O ^bgt,fst $dev
mount $dev $mnt -o clear_cache,nospace_cache
umount $mnt
btrfs ins dump-super $dev
...
compat_ro_flags 0x3
( FREE_SPACE_TREE |
FREE_SPACE_TREE_VALID )
...
This means a different behavior compared to bs >= ps cases.
[CAUSE]
The forcing usage of v2 space cache is done inside
btrfs_set_free_space_cache_settings(), however it never checks if we're
even using space cache but always enabling v2 cache.
[FIX]
Instead unconditionally enable v2 cache, only forcing v2 cache if the
old v1 cache is required.
Now v2 space cache can be properly disabled on bs < ps cases:
mkfs.btrfs -f -O ^bgt,fst $dev
mount $dev $mnt -o clear_cache,nospace_cache
umount $mnt
btrfs ins dump-super $dev
...
compat_ro_flags 0x0
...
Fixes: 9f73f1aef98b ("btrfs: force v2 space cache usage for subpage mount")
Reviewed-by: Filipe Manana <fdmanana@suse.com>
Signed-off-by: Qu Wenruo <wqu@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2025-12-18 15:15:28 +10:30
|
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|
2017-12-14 17:25:28 +09:00
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2023-11-22 12:17:50 -05:00
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2017-12-14 17:25:28 +09:00
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2023-11-22 12:17:51 -05:00
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2017-12-14 17:25:28 +09:00
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2023-11-22 12:17:51 -05:00
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2008-06-10 10:40:29 -04:00
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2023-11-22 12:17:51 -05:00
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2008-06-10 10:40:29 -04:00
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2023-11-22 12:17:51 -05:00
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2007-08-29 09:11:44 -04:00
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2020-02-21 14:56:12 +01:00
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2015-05-18 02:16:30 -07:00
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2020-02-06 10:24:26 -05:00
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2015-05-18 02:16:30 -07:00
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2025-11-18 17:06:46 +01:00
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2015-05-18 02:16:30 -07:00
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2025-11-18 17:06:46 +01:00
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2015-05-18 02:16:30 -07:00
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2017-06-22 02:26:54 +02:00
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2015-05-18 02:16:30 -07:00
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2021-07-29 05:22:16 -03:00
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2015-05-18 02:16:30 -07:00
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2021-07-29 05:22:16 -03:00
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2015-05-18 02:16:30 -07:00
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2020-05-15 19:35:55 +02:00
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2015-05-18 02:16:30 -07:00
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2020-02-06 10:24:26 -05:00
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2015-05-18 02:16:30 -07:00
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2021-07-29 05:22:16 -03:00
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2015-05-18 02:16:30 -07:00
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2021-07-29 05:22:16 -03:00
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2015-05-18 02:16:30 -07:00
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2020-01-24 09:33:01 -05:00
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2020-02-06 10:24:26 -05:00
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2015-05-18 02:16:30 -07:00
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2020-01-24 09:33:01 -05:00
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2015-05-18 02:16:30 -07:00
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Btrfs: change how we mount subvolumes
This work is in preperation for being able to set a different root as the
default mounting root.
There is currently a problem with how we mount subvolumes. We cannot currently
mount a subvolume of a subvolume, you can only mount subvolumes/snapshots of the
default subvolume. So say you take a snapshot of the default subvolume and call
it snap1, and then take a snapshot of snap1 and call it snap2, so now you have
/
/snap1
/snap1/snap2
as your available volumes. Currently you can only mount / and /snap1,
you cannot mount /snap1/snap2. To fix this problem instead of passing
subvolid=<name> you must pass in subvolid=<treeid>, where <treeid> is
the tree id that gets spit out via the subvolume listing you get from
the subvolume listing patches (btrfs filesystem list). This allows us
to mount /, /snap1 and /snap1/snap2 as the root volume.
In addition to the above, we also now read the default dir item in the
tree root to get the root key that it points to. For now this just
points at what has always been the default subvolme, but later on I plan
to change it to point at whatever root you want to be the new default
root, so you can just set the default mount and not have to mount with
-o subvolid=<treeid>. I tested this out with the above scenario and it
worked perfectly. Thanks,
mount -o subvol operates inside the selected subvolid. For example:
mount -o subvol=snap1,subvolid=256 /dev/xxx /mnt
/mnt will have the snap1 directory for the subvolume with id
256.
mount -o subvol=snap /dev/xxx /mnt
/mnt will be the snap directory of whatever the default subvolume
is.
Signed-off-by: Josef Bacik <josef@redhat.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2009-12-04 17:38:27 +00:00
|
|
|
|
2011-11-17 15:40:49 -05:00
|
|
|
|
Btrfs: change how we mount subvolumes
This work is in preperation for being able to set a different root as the
default mounting root.
There is currently a problem with how we mount subvolumes. We cannot currently
mount a subvolume of a subvolume, you can only mount subvolumes/snapshots of the
default subvolume. So say you take a snapshot of the default subvolume and call
it snap1, and then take a snapshot of snap1 and call it snap2, so now you have
/
/snap1
/snap1/snap2
as your available volumes. Currently you can only mount / and /snap1,
you cannot mount /snap1/snap2. To fix this problem instead of passing
subvolid=<name> you must pass in subvolid=<treeid>, where <treeid> is
the tree id that gets spit out via the subvolume listing you get from
the subvolume listing patches (btrfs filesystem list). This allows us
to mount /, /snap1 and /snap1/snap2 as the root volume.
In addition to the above, we also now read the default dir item in the
tree root to get the root key that it points to. For now this just
points at what has always been the default subvolme, but later on I plan
to change it to point at whatever root you want to be the new default
root, so you can just set the default mount and not have to mount with
-o subvolid=<treeid>. I tested this out with the above scenario and it
worked perfectly. Thanks,
mount -o subvol operates inside the selected subvolid. For example:
mount -o subvol=snap1,subvolid=256 /dev/xxx /mnt
/mnt will have the snap1 directory for the subvolume with id
256.
mount -o subvol=snap /dev/xxx /mnt
/mnt will be the snap directory of whatever the default subvolume
is.
Signed-off-by: Josef Bacik <josef@redhat.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2009-12-04 17:38:27 +00:00
|
|
|
|
2025-09-22 19:30:07 +08:00
|
|
|
|
Btrfs: change how we mount subvolumes
This work is in preperation for being able to set a different root as the
default mounting root.
There is currently a problem with how we mount subvolumes. We cannot currently
mount a subvolume of a subvolume, you can only mount subvolumes/snapshots of the
default subvolume. So say you take a snapshot of the default subvolume and call
it snap1, and then take a snapshot of snap1 and call it snap2, so now you have
/
/snap1
/snap1/snap2
as your available volumes. Currently you can only mount / and /snap1,
you cannot mount /snap1/snap2. To fix this problem instead of passing
subvolid=<name> you must pass in subvolid=<treeid>, where <treeid> is
the tree id that gets spit out via the subvolume listing you get from
the subvolume listing patches (btrfs filesystem list). This allows us
to mount /, /snap1 and /snap1/snap2 as the root volume.
In addition to the above, we also now read the default dir item in the
tree root to get the root key that it points to. For now this just
points at what has always been the default subvolme, but later on I plan
to change it to point at whatever root you want to be the new default
root, so you can just set the default mount and not have to mount with
-o subvolid=<treeid>. I tested this out with the above scenario and it
worked perfectly. Thanks,
mount -o subvol operates inside the selected subvolid. For example:
mount -o subvol=snap1,subvolid=256 /dev/xxx /mnt
/mnt will have the snap1 directory for the subvolume with id
256.
mount -o subvol=snap /dev/xxx /mnt
/mnt will be the snap directory of whatever the default subvolume
is.
Signed-off-by: Josef Bacik <josef@redhat.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2009-12-04 17:38:27 +00:00
|
|
|
|
2022-10-20 12:58:27 -04:00
|
|
|
|
Btrfs: change how we mount subvolumes
This work is in preperation for being able to set a different root as the
default mounting root.
There is currently a problem with how we mount subvolumes. We cannot currently
mount a subvolume of a subvolume, you can only mount subvolumes/snapshots of the
default subvolume. So say you take a snapshot of the default subvolume and call
it snap1, and then take a snapshot of snap1 and call it snap2, so now you have
/
/snap1
/snap1/snap2
as your available volumes. Currently you can only mount / and /snap1,
you cannot mount /snap1/snap2. To fix this problem instead of passing
subvolid=<name> you must pass in subvolid=<treeid>, where <treeid> is
the tree id that gets spit out via the subvolume listing you get from
the subvolume listing patches (btrfs filesystem list). This allows us
to mount /, /snap1 and /snap1/snap2 as the root volume.
In addition to the above, we also now read the default dir item in the
tree root to get the root key that it points to. For now this just
points at what has always been the default subvolme, but later on I plan
to change it to point at whatever root you want to be the new default
root, so you can just set the default mount and not have to mount with
-o subvolid=<treeid>. I tested this out with the above scenario and it
worked perfectly. Thanks,
mount -o subvol operates inside the selected subvolid. For example:
mount -o subvol=snap1,subvolid=256 /dev/xxx /mnt
/mnt will have the snap1 directory for the subvolume with id
256.
mount -o subvol=snap /dev/xxx /mnt
/mnt will be the snap directory of whatever the default subvolume
is.
Signed-off-by: Josef Bacik <josef@redhat.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2009-12-04 17:38:27 +00:00
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
2015-05-18 02:16:30 -07:00
|
|
|
|
Btrfs: change how we mount subvolumes
This work is in preperation for being able to set a different root as the
default mounting root.
There is currently a problem with how we mount subvolumes. We cannot currently
mount a subvolume of a subvolume, you can only mount subvolumes/snapshots of the
default subvolume. So say you take a snapshot of the default subvolume and call
it snap1, and then take a snapshot of snap1 and call it snap2, so now you have
/
/snap1
/snap1/snap2
as your available volumes. Currently you can only mount / and /snap1,
you cannot mount /snap1/snap2. To fix this problem instead of passing
subvolid=<name> you must pass in subvolid=<treeid>, where <treeid> is
the tree id that gets spit out via the subvolume listing you get from
the subvolume listing patches (btrfs filesystem list). This allows us
to mount /, /snap1 and /snap1/snap2 as the root volume.
In addition to the above, we also now read the default dir item in the
tree root to get the root key that it points to. For now this just
points at what has always been the default subvolme, but later on I plan
to change it to point at whatever root you want to be the new default
root, so you can just set the default mount and not have to mount with
-o subvolid=<treeid>. I tested this out with the above scenario and it
worked perfectly. Thanks,
mount -o subvol operates inside the selected subvolid. For example:
mount -o subvol=snap1,subvolid=256 /dev/xxx /mnt
/mnt will have the snap1 directory for the subvolume with id
256.
mount -o subvol=snap /dev/xxx /mnt
/mnt will be the snap directory of whatever the default subvolume
is.
Signed-off-by: Josef Bacik <josef@redhat.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2009-12-04 17:38:27 +00:00
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
2011-11-17 15:40:49 -05:00
|
|
|
|
2022-10-20 12:58:25 -04:00
|
|
|
|
2011-05-14 07:10:51 +00:00
|
|
|
|
2015-05-18 02:16:30 -07:00
|
|
|
|
2011-05-14 07:10:51 +00:00
|
|
|
|
Btrfs: change how we mount subvolumes
This work is in preperation for being able to set a different root as the
default mounting root.
There is currently a problem with how we mount subvolumes. We cannot currently
mount a subvolume of a subvolume, you can only mount subvolumes/snapshots of the
default subvolume. So say you take a snapshot of the default subvolume and call
it snap1, and then take a snapshot of snap1 and call it snap2, so now you have
/
/snap1
/snap1/snap2
as your available volumes. Currently you can only mount / and /snap1,
you cannot mount /snap1/snap2. To fix this problem instead of passing
subvolid=<name> you must pass in subvolid=<treeid>, where <treeid> is
the tree id that gets spit out via the subvolume listing you get from
the subvolume listing patches (btrfs filesystem list). This allows us
to mount /, /snap1 and /snap1/snap2 as the root volume.
In addition to the above, we also now read the default dir item in the
tree root to get the root key that it points to. For now this just
points at what has always been the default subvolme, but later on I plan
to change it to point at whatever root you want to be the new default
root, so you can just set the default mount and not have to mount with
-o subvolid=<treeid>. I tested this out with the above scenario and it
worked perfectly. Thanks,
mount -o subvol operates inside the selected subvolid. For example:
mount -o subvol=snap1,subvolid=256 /dev/xxx /mnt
/mnt will have the snap1 directory for the subvolume with id
256.
mount -o subvol=snap /dev/xxx /mnt
/mnt will be the snap directory of whatever the default subvolume
is.
Signed-off-by: Josef Bacik <josef@redhat.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2009-12-04 17:38:27 +00:00
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
2015-05-18 02:16:30 -07:00
|
|
|
|
Btrfs: change how we mount subvolumes
This work is in preperation for being able to set a different root as the
default mounting root.
There is currently a problem with how we mount subvolumes. We cannot currently
mount a subvolume of a subvolume, you can only mount subvolumes/snapshots of the
default subvolume. So say you take a snapshot of the default subvolume and call
it snap1, and then take a snapshot of snap1 and call it snap2, so now you have
/
/snap1
/snap1/snap2
as your available volumes. Currently you can only mount / and /snap1,
you cannot mount /snap1/snap2. To fix this problem instead of passing
subvolid=<name> you must pass in subvolid=<treeid>, where <treeid> is
the tree id that gets spit out via the subvolume listing you get from
the subvolume listing patches (btrfs filesystem list). This allows us
to mount /, /snap1 and /snap1/snap2 as the root volume.
In addition to the above, we also now read the default dir item in the
tree root to get the root key that it points to. For now this just
points at what has always been the default subvolme, but later on I plan
to change it to point at whatever root you want to be the new default
root, so you can just set the default mount and not have to mount with
-o subvolid=<treeid>. I tested this out with the above scenario and it
worked perfectly. Thanks,
mount -o subvol operates inside the selected subvolid. For example:
mount -o subvol=snap1,subvolid=256 /dev/xxx /mnt
/mnt will have the snap1 directory for the subvolume with id
256.
mount -o subvol=snap /dev/xxx /mnt
/mnt will be the snap directory of whatever the default subvolume
is.
Signed-off-by: Josef Bacik <josef@redhat.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2009-12-04 17:38:27 +00:00
|
|
|
|
2015-05-18 02:16:30 -07:00
|
|
|
|
|
|
|
|
|
Btrfs: change how we mount subvolumes
This work is in preperation for being able to set a different root as the
default mounting root.
There is currently a problem with how we mount subvolumes. We cannot currently
mount a subvolume of a subvolume, you can only mount subvolumes/snapshots of the
default subvolume. So say you take a snapshot of the default subvolume and call
it snap1, and then take a snapshot of snap1 and call it snap2, so now you have
/
/snap1
/snap1/snap2
as your available volumes. Currently you can only mount / and /snap1,
you cannot mount /snap1/snap2. To fix this problem instead of passing
subvolid=<name> you must pass in subvolid=<treeid>, where <treeid> is
the tree id that gets spit out via the subvolume listing you get from
the subvolume listing patches (btrfs filesystem list). This allows us
to mount /, /snap1 and /snap1/snap2 as the root volume.
In addition to the above, we also now read the default dir item in the
tree root to get the root key that it points to. For now this just
points at what has always been the default subvolme, but later on I plan
to change it to point at whatever root you want to be the new default
root, so you can just set the default mount and not have to mount with
-o subvolid=<treeid>. I tested this out with the above scenario and it
worked perfectly. Thanks,
mount -o subvol operates inside the selected subvolid. For example:
mount -o subvol=snap1,subvolid=256 /dev/xxx /mnt
/mnt will have the snap1 directory for the subvolume with id
256.
mount -o subvol=snap /dev/xxx /mnt
/mnt will be the snap directory of whatever the default subvolume
is.
Signed-off-by: Josef Bacik <josef@redhat.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2009-12-04 17:38:27 +00:00
|
|
|
|
|
|
|
|
|
|
|
|
|
|
2015-05-18 02:16:30 -07:00
|
|
|
|
|
|
|
|
|
Btrfs: change how we mount subvolumes
This work is in preperation for being able to set a different root as the
default mounting root.
There is currently a problem with how we mount subvolumes. We cannot currently
mount a subvolume of a subvolume, you can only mount subvolumes/snapshots of the
default subvolume. So say you take a snapshot of the default subvolume and call
it snap1, and then take a snapshot of snap1 and call it snap2, so now you have
/
/snap1
/snap1/snap2
as your available volumes. Currently you can only mount / and /snap1,
you cannot mount /snap1/snap2. To fix this problem instead of passing
subvolid=<name> you must pass in subvolid=<treeid>, where <treeid> is
the tree id that gets spit out via the subvolume listing you get from
the subvolume listing patches (btrfs filesystem list). This allows us
to mount /, /snap1 and /snap1/snap2 as the root volume.
In addition to the above, we also now read the default dir item in the
tree root to get the root key that it points to. For now this just
points at what has always been the default subvolme, but later on I plan
to change it to point at whatever root you want to be the new default
root, so you can just set the default mount and not have to mount with
-o subvolid=<treeid>. I tested this out with the above scenario and it
worked perfectly. Thanks,
mount -o subvol operates inside the selected subvolid. For example:
mount -o subvol=snap1,subvolid=256 /dev/xxx /mnt
/mnt will have the snap1 directory for the subvolume with id
256.
mount -o subvol=snap /dev/xxx /mnt
/mnt will be the snap directory of whatever the default subvolume
is.
Signed-off-by: Josef Bacik <josef@redhat.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2009-12-04 17:38:27 +00:00
|
|
|
|
|
|
|
|
|
2009-01-05 21:25:51 -05:00
|
|
|
|
2024-10-09 16:32:17 +02:00
|
|
|
|
2007-03-21 11:12:56 -04:00
|
|
|
|
2025-03-07 12:16:10 +00:00
|
|
|
|
2011-11-17 15:40:49 -05:00
|
|
|
|
2025-05-30 18:18:46 +02:00
|
|
|
|
2007-03-21 11:12:56 -04:00
|
|
|
|
|
|
|
|
|
|
|
|
|
|
2007-03-22 12:13:20 -04:00
|
|
|
|
2025-02-23 19:39:47 -05:00
|
|
|
|
2008-07-21 02:01:56 +05:30
|
|
|
|
2021-06-30 13:01:49 -07:00
|
|
|
|
|
|
|
|
|
|
|
|
|
|
2007-11-16 11:45:54 -05:00
|
|
|
|
2007-03-21 11:12:56 -04:00
|
|
|
|
2024-11-15 10:30:32 -05:00
|
|
|
|
2017-04-12 12:24:32 +02:00
|
|
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|
2025-05-30 18:18:46 +02:00
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2017-04-12 12:24:32 +02:00
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2025-05-30 18:18:46 +02:00
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2017-04-12 12:24:32 +02:00
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2025-05-30 18:18:46 +02:00
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2007-03-22 12:13:20 -04:00
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2007-08-27 16:49:44 -04:00
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2025-08-13 03:00:06 +09:00
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2024-06-13 11:36:26 +01:00
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Btrfs: Mixed back reference (FORWARD ROLLING FORMAT CHANGE)
This commit introduces a new kind of back reference for btrfs metadata.
Once a filesystem has been mounted with this commit, IT WILL NO LONGER
BE MOUNTABLE BY OLDER KERNELS.
When a tree block in subvolume tree is cow'd, the reference counts of all
extents it points to are increased by one. At transaction commit time,
the old root of the subvolume is recorded in a "dead root" data structure,
and the btree it points to is later walked, dropping reference counts
and freeing any blocks where the reference count goes to 0.
The increments done during cow and decrements done after commit cancel out,
and the walk is a very expensive way to go about freeing the blocks that
are no longer referenced by the new btree root. This commit reduces the
transaction overhead by avoiding the need for dead root records.
When a non-shared tree block is cow'd, we free the old block at once, and the
new block inherits old block's references. When a tree block with reference
count > 1 is cow'd, we increase the reference counts of all extents
the new block points to by one, and decrease the old block's reference count by
one.
This dead tree avoidance code removes the need to modify the reference
counts of lower level extents when a non-shared tree block is cow'd.
But we still need to update back ref for all pointers in the block.
This is because the location of the block is recorded in the back ref
item.
We can solve this by introducing a new type of back ref. The new
back ref provides information about pointer's key, level and in which
tree the pointer lives. This information allow us to find the pointer
by searching the tree. The shortcoming of the new back ref is that it
only works for pointers in tree blocks referenced by their owner trees.
This is mostly a problem for snapshots, where resolving one of these
fuzzy back references would be O(number_of_snapshots) and quite slow.
The solution used here is to use the fuzzy back references in the common
case where a given tree block is only referenced by one root,
and use the full back references when multiple roots have a reference
on a given block.
This commit adds per subvolume red-black tree to keep trace of cached
inodes. The red-black tree helps the balancing code to find cached
inodes whose inode numbers within a given range.
This commit improves the balancing code by introducing several data
structures to keep the state of balancing. The most important one
is the back ref cache. It caches how the upper level tree blocks are
referenced. This greatly reduce the overhead of checking back ref.
The improved balancing code scales significantly better with a large
number of snapshots.
This is a very large commit and was written in a number of
pieces. But, they depend heavily on the disk format change and were
squashed together to make sure git bisect didn't end up in a
bad state wrt space balancing or the format change.
Signed-off-by: Yan Zheng <zheng.yan@oracle.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2009-06-10 10:45:14 -04:00
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2025-05-30 18:18:46 +02:00
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2007-06-12 06:35:45 -04:00
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2007-03-21 11:12:56 -04:00
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2025-03-07 12:16:10 +00:00
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2012-01-08 22:15:13 -05:00
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2025-05-30 18:18:46 +02:00
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2007-06-12 06:35:45 -04:00
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2007-03-21 11:12:56 -04:00
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2007-08-29 15:47:34 -04:00
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2017-11-27 13:05:09 -08:00
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2007-03-21 11:12:56 -04:00
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2007-06-12 06:35:45 -04:00
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2016-06-21 21:16:51 -04:00
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2025-05-30 18:18:46 +02:00
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2007-03-23 10:01:08 -04:00
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2008-06-10 10:07:39 -04:00
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2007-03-23 10:01:08 -04:00
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2011-11-17 15:40:49 -05:00
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2007-03-23 11:00:45 -04:00
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2016-06-09 17:27:55 -04:00
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Btrfs: add initial tracepoint support for btrfs
Tracepoints can provide insight into why btrfs hits bugs and be greatly
helpful for debugging, e.g
dd-7822 [000] 2121.641088: btrfs_inode_request: root = 5(FS_TREE), gen = 4, ino = 256, blocks = 8, disk_i_size = 0, last_trans = 8, logged_trans = 0
dd-7822 [000] 2121.641100: btrfs_inode_new: root = 5(FS_TREE), gen = 8, ino = 257, blocks = 0, disk_i_size = 0, last_trans = 0, logged_trans = 0
btrfs-transacti-7804 [001] 2146.935420: btrfs_cow_block: root = 2(EXTENT_TREE), refs = 2, orig_buf = 29368320 (orig_level = 0), cow_buf = 29388800 (cow_level = 0)
btrfs-transacti-7804 [001] 2146.935473: btrfs_cow_block: root = 1(ROOT_TREE), refs = 2, orig_buf = 29364224 (orig_level = 0), cow_buf = 29392896 (cow_level = 0)
btrfs-transacti-7804 [001] 2146.972221: btrfs_transaction_commit: root = 1(ROOT_TREE), gen = 8
flush-btrfs-2-7821 [001] 2155.824210: btrfs_chunk_alloc: root = 3(CHUNK_TREE), offset = 1103101952, size = 1073741824, num_stripes = 1, sub_stripes = 0, type = DATA
flush-btrfs-2-7821 [001] 2155.824241: btrfs_cow_block: root = 2(EXTENT_TREE), refs = 2, orig_buf = 29388800 (orig_level = 0), cow_buf = 29396992 (cow_level = 0)
flush-btrfs-2-7821 [001] 2155.824255: btrfs_cow_block: root = 4(DEV_TREE), refs = 2, orig_buf = 29372416 (orig_level = 0), cow_buf = 29401088 (cow_level = 0)
flush-btrfs-2-7821 [000] 2155.824329: btrfs_cow_block: root = 3(CHUNK_TREE), refs = 2, orig_buf = 20971520 (orig_level = 0), cow_buf = 20975616 (cow_level = 0)
btrfs-endio-wri-7800 [001] 2155.898019: btrfs_cow_block: root = 5(FS_TREE), refs = 2, orig_buf = 29384704 (orig_level = 0), cow_buf = 29405184 (cow_level = 0)
btrfs-endio-wri-7800 [001] 2155.898043: btrfs_cow_block: root = 7(CSUM_TREE), refs = 2, orig_buf = 29376512 (orig_level = 0), cow_buf = 29409280 (cow_level = 0)
Here is what I have added:
1) ordere_extent:
btrfs_ordered_extent_add
btrfs_ordered_extent_remove
btrfs_ordered_extent_start
btrfs_ordered_extent_put
These provide critical information to understand how ordered_extents are
updated.
2) extent_map:
btrfs_get_extent
extent_map is used in both read and write cases, and it is useful for tracking
how btrfs specific IO is running.
3) writepage:
__extent_writepage
btrfs_writepage_end_io_hook
Pages are cirtical resourses and produce a lot of corner cases during writeback,
so it is valuable to know how page is written to disk.
4) inode:
btrfs_inode_new
btrfs_inode_request
btrfs_inode_evict
These can show where and when a inode is created, when a inode is evicted.
5) sync:
btrfs_sync_file
btrfs_sync_fs
These show sync arguments.
6) transaction:
btrfs_transaction_commit
In transaction based filesystem, it will be useful to know the generation and
who does commit.
7) back reference and cow:
btrfs_delayed_tree_ref
btrfs_delayed_data_ref
btrfs_delayed_ref_head
btrfs_cow_block
Btrfs natively supports back references, these tracepoints are helpful on
understanding btrfs's COW mechanism.
8) chunk:
btrfs_chunk_alloc
btrfs_chunk_free
Chunk is a link between physical offset and logical offset, and stands for space
infomation in btrfs, and these are helpful on tracing space things.
9) reserved_extent:
btrfs_reserved_extent_alloc
btrfs_reserved_extent_free
These can show how btrfs uses its space.
Signed-off-by: Liu Bo <liubo2009@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-03-24 11:18:59 +00:00
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2007-03-23 19:47:49 -04:00
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2011-11-17 15:40:49 -05:00
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2007-03-23 19:47:49 -04:00
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2008-11-06 22:02:51 -05:00
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2024-05-14 16:48:12 +02:00
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2008-11-06 22:02:51 -05:00
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Btrfs: fix uncompleted transaction
In some cases, we need commit the current transaction, but don't want
to start a new one if there is no running transaction, so we introduce
the function - btrfs_attach_transaction(), which can catch the current
transaction, and return -ENOENT if there is no running transaction.
But no running transaction doesn't mean the current transction completely,
because we removed the running transaction before it completes. In some
cases, it doesn't matter. But in some special cases, such as freeze fs, we
hope the transaction is fully on disk, it will introduce some bugs, for
example, we may feeze the fs and dump the data in the disk, if the transction
doesn't complete, we would dump inconsistent data. So we need fix the above
problem for those cases.
We fixes this problem by introducing a function:
btrfs_attach_transaction_barrier()
if we hope all the transaction is fully on the disk, even they are not
running, we can use this function.
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Josef Bacik <jbacik@fusionio.com>
2013-02-20 09:17:06 +00:00
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2012-09-14 10:34:40 -04:00
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Btrfs: fix orphan transaction on the freezed filesystem
With the following debug patch:
static int btrfs_freeze(struct super_block *sb)
{
+ struct btrfs_fs_info *fs_info = btrfs_sb(sb);
+ struct btrfs_transaction *trans;
+
+ spin_lock(&fs_info->trans_lock);
+ trans = fs_info->running_transaction;
+ if (trans) {
+ printk("Transid %llu, use_count %d, num_writer %d\n",
+ trans->transid, atomic_read(&trans->use_count),
+ atomic_read(&trans->num_writers));
+ }
+ spin_unlock(&fs_info->trans_lock);
return 0;
}
I found there was a orphan transaction after the freeze operation was done.
It is because the transaction may not be committed when the transaction handle
end even though it is the last handle of the current transaction. This design
avoid committing the transaction frequently, but also introduce the above
problem.
So I add btrfs_attach_transaction() which can catch the current transaction
and commit it. If there is no transaction, it will return ENOENT, and do not
anything.
This function also can be used to instead of btrfs_join_transaction_freeze()
because it don't increase the writer counter and don't start a new transaction,
so it also can fix the deadlock between sync and freeze.
Besides that, it is used to instead of btrfs_join_transaction() in
transaction_kthread(), because if there is no transaction, the transaction
kthread needn't anything.
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
2012-09-20 01:54:00 -06:00
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2014-03-28 17:38:48 +01:00
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2022-10-19 10:50:55 -04:00
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2014-03-28 17:38:48 +01:00
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2015-01-19 15:42:41 +08:00
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2017-10-10 13:48:05 +03:00
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2015-01-19 15:42:41 +08:00
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2014-03-28 17:38:48 +01:00
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2015-01-19 14:21:02 +01:00
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2012-09-14 10:34:40 -04:00
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2016-09-09 21:39:03 -04:00
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2007-03-23 10:01:08 -04:00
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2020-10-16 11:29:16 -04:00
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2011-12-08 21:32:45 -05:00
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2009-04-02 16:46:06 -04:00
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2011-11-17 15:40:49 -05:00
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2017-10-31 18:06:34 +01:00
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btrfs: don't show full path of bind mounts in subvol=
Chris Murphy reported a problem where rpm ostree will bind mount a bunch
of things for whatever voodoo it's doing. But when it does this
/proc/mounts shows something like
/dev/sda /mnt/test btrfs rw,relatime,subvolid=256,subvol=/foo 0 0
/dev/sda /mnt/test/baz btrfs rw,relatime,subvolid=256,subvol=/foo/bar 0 0
Despite subvolid=256 being subvol=/foo. This is because we're just
spitting out the dentry of the mount point, which in the case of bind
mounts is the source path for the mountpoint. Instead we should spit
out the path to the actual subvol. Fix this by looking up the name for
the subvolid we have mounted. With this fix the same test looks like
this
/dev/sda /mnt/test btrfs rw,relatime,subvolid=256,subvol=/foo 0 0
/dev/sda /mnt/test/baz btrfs rw,relatime,subvolid=256,subvol=/foo 0 0
Reported-by: Chris Murphy <chris@colorremedies.com>
CC: stable@vger.kernel.org # 4.4+
Signed-off-by: Josef Bacik <josef@toxicpanda.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2020-07-22 11:12:46 -04:00
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2020-10-16 11:29:16 -04:00
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2009-04-02 16:46:06 -04:00
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2016-06-09 21:38:35 -04:00
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2009-04-02 16:46:06 -04:00
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2016-06-09 21:38:35 -04:00
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2009-04-02 16:46:06 -04:00
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2016-06-09 21:38:35 -04:00
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2009-04-02 16:46:06 -04:00
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2016-06-09 21:38:35 -04:00
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2009-04-02 16:46:06 -04:00
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2013-08-08 22:45:48 +01:00
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2013-08-20 13:20:07 +02:00
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2009-04-02 16:46:06 -04:00
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2018-02-13 17:50:42 +08:00
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2016-06-09 21:38:35 -04:00
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2017-10-31 18:06:34 +01:00
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2016-06-09 21:38:35 -04:00
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2011-03-31 00:44:29 +00:00
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btrfs: accept and ignore compression level for lzo
The compression level is meaningless for lzo, but before commit
3f093ccb95f30 ("btrfs: harden parsing of compression mount options"),
it was silently ignored if passed.
After that commit, passing a level with lzo fails to mount:
BTRFS error: unrecognized compression value lzo:1
It seems reasonable for users to expect that lzo would permit a numeric
level option, as all the other algos do, even though the kernel's
implementation of LZO currently only supports a single level. Because it
has always worked to pass a level, it seems likely to me that users in
the real world are relying on doing so.
This patch restores the old behavior, giving "lzo:N" the same semantics
as all of the other compression algos.
To be clear, silly variants like "lzo:one", "lzo:the_first_option", or
"lzo:armageddon" also used to work. This isn't meant to suggest that
any possible mis-interpretation of mount options that once worked must
continue to work forever. This is an exceptional case where it makes
sense to preserve compatibility, both because the mis-interpretation is
reasonable, and because nothing tangible is sacrificed.
Finally update btrfs_show_options() to ignore the level of LZO, as it
is only the default level without any extra meaning.
Fixes: 3f093ccb95f30 ("btrfs: harden parsing of compression mount options")
Reviewed-by: Daniel Vacek <neelx@suse.com>
Reviewed-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: Calvin Owens <calvin@wbinvd.org>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2025-08-25 18:32:04 +09:30
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2017-09-15 17:36:58 +02:00
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2011-03-31 00:44:29 +00:00
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2016-06-09 21:38:35 -04:00
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2009-06-10 09:51:32 -04:00
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2016-06-09 21:38:35 -04:00
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2009-06-09 20:28:34 -04:00
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2016-06-09 21:38:35 -04:00
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2009-04-02 16:46:06 -04:00
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2016-06-09 21:38:35 -04:00
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2009-05-14 13:52:21 -04:00
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2016-06-09 21:38:35 -04:00
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2020-10-16 11:29:16 -04:00
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2020-10-16 11:29:17 -04:00
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2020-10-16 11:29:18 -04:00
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2020-10-16 11:29:19 -04:00
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2024-06-14 13:52:30 +09:30
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2024-06-14 13:52:31 +09:30
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2016-06-09 21:38:35 -04:00
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2009-05-14 13:52:21 -04:00
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2019-12-13 16:22:11 -08:00
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2009-12-14 22:01:12 +00:00
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2019-12-13 16:22:14 -08:00
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2017-11-27 13:05:09 -08:00
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2009-04-02 16:46:06 -04:00
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2020-11-18 15:06:23 -08:00
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2011-03-31 00:44:29 +00:00
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2020-11-18 15:06:23 -08:00
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2015-09-29 20:50:38 -07:00
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2011-10-03 14:07:49 -04:00
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2011-11-11 10:14:57 -05:00
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2016-06-09 21:38:35 -04:00
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2013-08-15 17:11:24 +02:00
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2016-06-09 21:38:35 -04:00
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2011-03-31 00:44:29 +00:00
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2016-06-09 21:38:35 -04:00
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2011-03-31 00:44:29 +00:00
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2016-06-09 21:38:35 -04:00
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2011-06-28 15:10:37 +00:00
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2016-06-09 21:38:35 -04:00
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2011-06-28 15:10:37 +00:00
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2016-06-09 21:38:35 -04:00
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2012-01-16 22:04:48 +02:00
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2013-07-24 10:30:02 +08:00
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2018-02-13 17:50:44 +08:00
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2016-06-09 21:38:35 -04:00
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2011-10-03 23:22:31 -04:00
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2013-08-01 18:14:52 +02:00
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2018-02-13 17:50:46 +08:00
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2015-09-23 14:54:14 -04:00
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2016-06-09 21:38:35 -04:00
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2015-09-23 14:54:14 -04:00
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2016-06-09 21:38:35 -04:00
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2015-09-23 14:54:14 -04:00
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2017-09-29 15:43:48 -04:00
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2025-08-12 16:04:41 -07:00
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2024-04-15 16:16:23 -04:00
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btrfs: don't show full path of bind mounts in subvol=
Chris Murphy reported a problem where rpm ostree will bind mount a bunch
of things for whatever voodoo it's doing. But when it does this
/proc/mounts shows something like
/dev/sda /mnt/test btrfs rw,relatime,subvolid=256,subvol=/foo 0 0
/dev/sda /mnt/test/baz btrfs rw,relatime,subvolid=256,subvol=/foo/bar 0 0
Despite subvolid=256 being subvol=/foo. This is because we're just
spitting out the dentry of the mount point, which in the case of bind
mounts is the source path for the mountpoint. Instead we should spit
out the path to the actual subvol. Fix this by looking up the name for
the subvolid we have mounted. With this fix the same test looks like
this
/dev/sda /mnt/test btrfs rw,relatime,subvolid=256,subvol=/foo 0 0
/dev/sda /mnt/test/baz btrfs rw,relatime,subvolid=256,subvol=/foo 0 0
Reported-by: Chris Murphy <chris@colorremedies.com>
CC: stable@vger.kernel.org # 4.4+
Signed-off-by: Josef Bacik <josef@toxicpanda.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2020-07-22 11:12:46 -04:00
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2024-04-15 16:16:23 -04:00
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|
btrfs: don't show full path of bind mounts in subvol=
Chris Murphy reported a problem where rpm ostree will bind mount a bunch
of things for whatever voodoo it's doing. But when it does this
/proc/mounts shows something like
/dev/sda /mnt/test btrfs rw,relatime,subvolid=256,subvol=/foo 0 0
/dev/sda /mnt/test/baz btrfs rw,relatime,subvolid=256,subvol=/foo/bar 0 0
Despite subvolid=256 being subvol=/foo. This is because we're just
spitting out the dentry of the mount point, which in the case of bind
mounts is the source path for the mountpoint. Instead we should spit
out the path to the actual subvol. Fix this by looking up the name for
the subvolid we have mounted. With this fix the same test looks like
this
/dev/sda /mnt/test btrfs rw,relatime,subvolid=256,subvol=/foo 0 0
/dev/sda /mnt/test/baz btrfs rw,relatime,subvolid=256,subvol=/foo 0 0
Reported-by: Chris Murphy <chris@colorremedies.com>
CC: stable@vger.kernel.org # 4.4+
Signed-off-by: Josef Bacik <josef@toxicpanda.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2020-07-22 11:12:46 -04:00
|
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2025-03-19 22:49:00 +01:00
|
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|
btrfs: don't show full path of bind mounts in subvol=
Chris Murphy reported a problem where rpm ostree will bind mount a bunch
of things for whatever voodoo it's doing. But when it does this
/proc/mounts shows something like
/dev/sda /mnt/test btrfs rw,relatime,subvolid=256,subvol=/foo 0 0
/dev/sda /mnt/test/baz btrfs rw,relatime,subvolid=256,subvol=/foo/bar 0 0
Despite subvolid=256 being subvol=/foo. This is because we're just
spitting out the dentry of the mount point, which in the case of bind
mounts is the source path for the mountpoint. Instead we should spit
out the path to the actual subvol. Fix this by looking up the name for
the subvolid we have mounted. With this fix the same test looks like
this
/dev/sda /mnt/test btrfs rw,relatime,subvolid=256,subvol=/foo 0 0
/dev/sda /mnt/test/baz btrfs rw,relatime,subvolid=256,subvol=/foo 0 0
Reported-by: Chris Murphy <chris@colorremedies.com>
CC: stable@vger.kernel.org # 4.4+
Signed-off-by: Josef Bacik <josef@toxicpanda.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2020-07-22 11:12:46 -04:00
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2009-04-02 16:46:06 -04:00
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2011-09-29 13:11:33 +02:00
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2025-04-23 18:53:57 +02:00
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2011-09-29 13:11:33 +02:00
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2025-04-23 18:53:57 +02:00
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2011-09-29 13:11:33 +02:00
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2015-05-18 02:16:29 -07:00
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2019-03-29 14:03:17 +08:00
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2011-07-25 15:55:42 -04:00
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2015-05-18 02:16:28 -07:00
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2011-07-25 15:55:42 -04:00
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2015-05-18 02:16:30 -07:00
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2020-02-21 14:56:12 +01:00
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2015-05-18 02:16:30 -07:00
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2011-11-16 21:43:59 -05:00
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2015-05-18 02:16:28 -07:00
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2011-07-25 15:55:42 -04:00
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2015-05-18 02:16:29 -07:00
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2011-11-16 21:43:59 -05:00
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2016-09-20 10:05:02 -04:00
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2015-05-18 02:16:29 -07:00
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2024-04-15 16:16:23 -04:00
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2015-05-18 02:16:29 -07:00
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2016-09-20 10:05:02 -04:00
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2015-05-18 02:16:29 -07:00
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2015-05-18 02:16:30 -07:00
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2016-09-20 10:05:02 -04:00
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2015-05-18 02:16:29 -07:00
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2011-09-29 13:11:33 +02:00
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2015-05-18 02:16:28 -07:00
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2011-07-25 15:55:42 -04:00
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2010-11-19 14:59:15 -05:00
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2012-04-24 22:59:16 +03:00
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2018-02-13 17:50:42 +08:00
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2012-04-24 22:59:16 +03:00
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2013-12-20 11:37:06 -05:00
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2012-04-24 22:59:16 +03:00
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2014-02-28 10:46:06 +08:00
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2014-02-28 10:46:07 +08:00
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2014-02-28 10:46:12 +08:00
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2023-03-28 12:56:13 +09:00
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2014-02-28 10:46:10 +08:00
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2014-02-28 10:46:15 +08:00
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2012-04-24 22:59:16 +03:00
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2013-04-11 10:29:35 +00:00
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2024-07-13 18:33:44 +09:30
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2013-04-11 10:29:35 +00:00
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2013-02-20 23:32:52 -07:00
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2017-11-27 13:05:09 -08:00
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2013-02-20 23:32:52 -07:00
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2017-11-27 13:05:09 -08:00
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2013-02-20 23:32:52 -07:00
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2024-07-13 18:33:44 +09:30
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2013-02-20 23:32:52 -07:00
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btrfs: keep sb cache_generation consistent with space_cache
When mounting, btrfs uses the cache_generation in the super block to
determine if space cache v1 is in use. However, by mounting with
nospace_cache or space_cache=v2, it is possible to disable space cache
v1, which does not result in un-setting cache_generation back to 0.
In order to base some logic, like mount option printing in /proc/mounts,
on the current state of the space cache rather than just the values of
the mount option, keep the value of cache_generation consistent with the
status of space cache v1.
We ensure that cache_generation > 0 iff the file system is using
space_cache v1. This requires committing a transaction on any mount
which changes whether we are using v1. (v1->nospace_cache, v1->v2,
nospace_cache->v1, v2->v1).
Since the mechanism for writing out the cache generation is transaction
commit, but we want some finer grained control over when we un-set it,
we can't just rely on the SPACE_CACHE mount option, and introduce an
fs_info flag that mount can use when it wants to unset the generation.
Reviewed-by: Josef Bacik <josef@toxicpanda.com>
Signed-off-by: Boris Burkov <boris@bur.io>
Signed-off-by: David Sterba <dsterba@suse.com>
2020-11-18 15:06:22 -08:00
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2013-02-20 23:32:52 -07:00
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2025-08-22 00:57:42 +02:00
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2016-04-04 15:31:22 +01:00
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2013-02-20 23:32:52 -07:00
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2017-07-17 08:45:34 +01:00
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2013-02-20 23:32:52 -07:00
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2019-12-13 16:22:14 -08:00
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btrfs: keep sb cache_generation consistent with space_cache
When mounting, btrfs uses the cache_generation in the super block to
determine if space cache v1 is in use. However, by mounting with
nospace_cache or space_cache=v2, it is possible to disable space cache
v1, which does not result in un-setting cache_generation back to 0.
In order to base some logic, like mount option printing in /proc/mounts,
on the current state of the space cache rather than just the values of
the mount option, keep the value of cache_generation consistent with the
status of space cache v1.
We ensure that cache_generation > 0 iff the file system is using
space_cache v1. This requires committing a transaction on any mount
which changes whether we are using v1. (v1->nospace_cache, v1->v2,
nospace_cache->v1, v2->v1).
Since the mechanism for writing out the cache generation is transaction
commit, but we want some finer grained control over when we un-set it,
we can't just rely on the SPACE_CACHE mount option, and introduce an
fs_info flag that mount can use when it wants to unset the generation.
Reviewed-by: Josef Bacik <josef@toxicpanda.com>
Signed-off-by: Boris Burkov <boris@bur.io>
Signed-off-by: David Sterba <dsterba@suse.com>
2020-11-18 15:06:22 -08:00
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2013-02-20 23:32:52 -07:00
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2023-11-22 12:17:42 -05:00
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2008-11-12 14:34:12 -05:00
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2026-04-01 18:51:35 +01:00
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2023-11-22 12:17:42 -05:00
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2023-11-22 12:17:47 -05:00
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2025-08-13 03:00:07 +09:00
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2023-11-22 12:17:47 -05:00
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2024-06-14 13:52:30 +09:30
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2024-06-14 13:52:31 +09:30
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2023-11-22 12:17:47 -05:00
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2025-08-13 03:00:07 +09:00
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2023-11-22 12:17:47 -05:00
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2025-07-23 00:38:37 +09:00
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2023-11-22 12:17:47 -05:00
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2023-11-22 12:17:49 -05:00
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2023-11-22 12:17:47 -05:00
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2024-01-10 12:54:41 -05:00
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2014-03-13 10:14:33 -04:00
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2020-07-23 00:18:04 +09:00
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2013-02-20 23:32:52 -07:00
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btrfs: reject ro->rw reconfiguration if there are hard ro requirements
[BUG]
Syzbot reports the following crash:
BTRFS info (device loop0 state MCS): disabling free space tree
BTRFS info (device loop0 state MCS): clearing compat-ro feature flag for FREE_SPACE_TREE (0x1)
BTRFS info (device loop0 state MCS): clearing compat-ro feature flag for FREE_SPACE_TREE_VALID (0x2)
Oops: general protection fault, probably for non-canonical address 0xdffffc0000000003: 0000 [#1] PREEMPT SMP KASAN NOPTI
KASAN: null-ptr-deref in range [0x0000000000000018-0x000000000000001f]
Hardware name: QEMU Standard PC (Q35 + ICH9, 2009), BIOS 1.16.3-debian-1.16.3-2~bpo12+1 04/01/2014
RIP: 0010:backup_super_roots fs/btrfs/disk-io.c:1691 [inline]
RIP: 0010:write_all_supers+0x97a/0x40f0 fs/btrfs/disk-io.c:4041
Call Trace:
<TASK>
btrfs_commit_transaction+0x1eae/0x3740 fs/btrfs/transaction.c:2530
btrfs_delete_free_space_tree+0x383/0x730 fs/btrfs/free-space-tree.c:1312
btrfs_start_pre_rw_mount+0xf28/0x1300 fs/btrfs/disk-io.c:3012
btrfs_remount_rw fs/btrfs/super.c:1309 [inline]
btrfs_reconfigure+0xae6/0x2d40 fs/btrfs/super.c:1534
btrfs_reconfigure_for_mount fs/btrfs/super.c:2020 [inline]
btrfs_get_tree_subvol fs/btrfs/super.c:2079 [inline]
btrfs_get_tree+0x918/0x1920 fs/btrfs/super.c:2115
vfs_get_tree+0x90/0x2b0 fs/super.c:1800
do_new_mount+0x2be/0xb40 fs/namespace.c:3472
do_mount fs/namespace.c:3812 [inline]
__do_sys_mount fs/namespace.c:4020 [inline]
__se_sys_mount+0x2d6/0x3c0 fs/namespace.c:3997
do_syscall_x64 arch/x86/entry/common.c:52 [inline]
do_syscall_64+0xf3/0x230 arch/x86/entry/common.c:83
entry_SYSCALL_64_after_hwframe+0x77/0x7f
[CAUSE]
To support mounting different subvolume with different RO/RW flags for
the new mount APIs, btrfs introduced two workaround to support this feature:
- Skip mount option/feature checks if we are mounting a different
subvolume
- Reconfigure the fs to RW if the initial mount is RO
Combining these two, we can have the following sequence:
- Mount the fs ro,rescue=all,clear_cache,space_cache=v1
rescue=all will mark the fs as hard read-only, so no v2 cache clearing
will happen.
- Mount a subvolume rw of the same fs.
We go into btrfs_get_tree_subvol(), but fc_mount() returns EBUSY
because our new fc is RW, different from the original fs.
Now we enter btrfs_reconfigure_for_mount(), which switches the RO flag
first so that we can grab the existing fs_info.
Then we reconfigure the fs to RW.
- During reconfiguration, option/features check is skipped
This means we will restart the v2 cache clearing, and convert back to
v1 cache.
This will trigger fs writes, and since the original fs has "rescue=all"
option, it skips the csum tree read.
And eventually causing NULL pointer dereference in super block
writeback.
[FIX]
For reconfiguration caused by different subvolume RO/RW flags, ensure we
always run btrfs_check_options() to ensure we have proper hard RO
requirements met.
In fact the function btrfs_check_options() doesn't really do many
complex checks, but hard RO requirement and some feature dependency
checks, thus there is no special reason not to do the check for mount
reconfiguration.
Reported-by: syzbot+56360f93efa90ff15870@syzkaller.appspotmail.com
Link: https://lore.kernel.org/linux-btrfs/0000000000008c5d090621cb2770@google.com/
Fixes: f044b318675f ("btrfs: handle the ro->rw transition for mounting different subvolumes")
CC: stable@vger.kernel.org # 6.8+
Reviewed-by: Johannes Thumshirn <johannes.thumshirn@wdc.com>
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2024-09-19 20:18:11 +09:30
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2023-11-22 12:17:47 -05:00
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2014-09-23 13:40:08 +08:00
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2023-11-22 12:17:47 -05:00
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btrfs: relax block-group-tree feature dependency checks
[BUG]
When one user did a wrong attempt to clear block group tree, which can
not be done through mount option, by using "-o clear_cache,space_cache=v2",
it will cause the following error on a fs with block-group-tree feature:
BTRFS info (device dm-1): force clearing of disk cache
BTRFS info (device dm-1): using free space tree
BTRFS info (device dm-1): clearing free space tree
BTRFS info (device dm-1): clearing compat-ro feature flag for FREE_SPACE_TREE (0x1)
BTRFS info (device dm-1): clearing compat-ro feature flag for FREE_SPACE_TREE_VALID (0x2)
BTRFS error (device dm-1): block-group-tree feature requires fres-space-tree and no-holes
BTRFS error (device dm-1): super block corruption detected before writing it to disk
BTRFS: error (device dm-1) in write_all_supers:4318: errno=-117 Filesystem corrupted (unexpected superblock corruption detected)
BTRFS warning (device dm-1: state E): Skipping commit of aborted transaction.
[CAUSE]
Although the dependency for block-group-tree feature is just an
artificial one (to reduce test matrix), we put the dependency check into
btrfs_validate_super().
This is too strict, and during space cache clearing, we will have a
window where free space tree is cleared, and we need to commit the super
block.
In that window, we had block group tree without v2 cache, and triggered
the artificial dependency check.
This is not necessary at all, especially for such a soft dependency.
[FIX]
Introduce a new helper, btrfs_check_features(), to do all the runtime
limitation checks, including:
- Unsupported incompat flags check
- Unsupported compat RO flags check
- Setting missing incompat flags
- Artificial feature dependency checks
Currently only block group tree will rely on this.
- Subpage runtime check for v1 cache
With this helper, we can move quite some checks from
open_ctree()/btrfs_remount() into it, and just call it after
btrfs_parse_options().
Now "-o clear_cache,space_cache=v2" will not trigger the above error
anymore.
Signed-off-by: Qu Wenruo <wqu@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
[ edit messages ]
Signed-off-by: David Sterba <dsterba@suse.com>
2022-09-12 13:44:37 +08:00
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2023-11-22 12:17:47 -05:00
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|
btrfs: relax block-group-tree feature dependency checks
[BUG]
When one user did a wrong attempt to clear block group tree, which can
not be done through mount option, by using "-o clear_cache,space_cache=v2",
it will cause the following error on a fs with block-group-tree feature:
BTRFS info (device dm-1): force clearing of disk cache
BTRFS info (device dm-1): using free space tree
BTRFS info (device dm-1): clearing free space tree
BTRFS info (device dm-1): clearing compat-ro feature flag for FREE_SPACE_TREE (0x1)
BTRFS info (device dm-1): clearing compat-ro feature flag for FREE_SPACE_TREE_VALID (0x2)
BTRFS error (device dm-1): block-group-tree feature requires fres-space-tree and no-holes
BTRFS error (device dm-1): super block corruption detected before writing it to disk
BTRFS: error (device dm-1) in write_all_supers:4318: errno=-117 Filesystem corrupted (unexpected superblock corruption detected)
BTRFS warning (device dm-1: state E): Skipping commit of aborted transaction.
[CAUSE]
Although the dependency for block-group-tree feature is just an
artificial one (to reduce test matrix), we put the dependency check into
btrfs_validate_super().
This is too strict, and during space cache clearing, we will have a
window where free space tree is cleared, and we need to commit the super
block.
In that window, we had block group tree without v2 cache, and triggered
the artificial dependency check.
This is not necessary at all, especially for such a soft dependency.
[FIX]
Introduce a new helper, btrfs_check_features(), to do all the runtime
limitation checks, including:
- Unsupported incompat flags check
- Unsupported compat RO flags check
- Setting missing incompat flags
- Artificial feature dependency checks
Currently only block group tree will rely on this.
- Subpage runtime check for v1 cache
With this helper, we can move quite some checks from
open_ctree()/btrfs_remount() into it, and just call it after
btrfs_parse_options().
Now "-o clear_cache,space_cache=v2" will not trigger the above error
anymore.
Signed-off-by: Qu Wenruo <wqu@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
[ edit messages ]
Signed-off-by: David Sterba <dsterba@suse.com>
2022-09-12 13:44:37 +08:00
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2023-11-22 12:17:47 -05:00
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2012-04-24 22:59:16 +03:00
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btrfs: fix spurious free_space_tree remount warning
The intended logic of the check is to catch cases where the desired
free_space_tree setting doesn't match the mounted setting, and the
remount is anything but ro->rw. However, it makes the mistake of
checking equality on a masked integer (btrfs_test_opt) against a boolean
(btrfs_fs_compat_ro).
If you run the reproducer:
$ mount -o space_cache=v2 dev mnt
$ mount -o remount,ro mnt
you would expect no warning, because the remount is not attempting to
change the free space tree setting, but we do see the warning.
To fix this, add explicit bool type casts to the condition.
I tested a variety of transitions:
sudo mount -o space_cache=v2 /dev/vg0/lv0 mnt/lol
(fst enabled)
mount -o remount,ro mnt/lol
(no warning, no fst change)
sudo mount -o remount,rw,space_cache=v1,clear_cache
(no warning, ro->rw)
sudo mount -o remount,rw,space_cache=v2 mnt
(warning, rw->rw with change)
sudo mount -o remount,ro mnt
(no warning, no fst change)
sudo mount -o remount,rw,space_cache=v2 mnt
(no warning, no fst change)
Reported-by: Chris Murphy <lists@colorremedies.com>
CC: stable@vger.kernel.org # 5.11
Signed-off-by: Boris Burkov <boris@bur.io>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2021-02-23 10:22:32 -08:00
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2023-11-22 12:17:47 -05:00
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2020-11-18 15:06:24 -08:00
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2023-11-22 12:17:47 -05:00
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2020-11-18 15:06:24 -08:00
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2023-11-22 12:17:47 -05:00
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2008-11-12 14:34:12 -05:00
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2020-07-30 11:18:09 -04:00
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2023-11-22 12:17:47 -05:00
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2020-07-30 11:18:09 -04:00
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2023-11-22 12:17:47 -05:00
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2020-07-30 11:18:09 -04:00
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2023-11-22 12:17:47 -05:00
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2014-02-20 08:48:07 -08:00
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2023-11-22 12:17:47 -05:00
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2020-11-18 15:06:20 -08:00
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2020-07-23 00:18:04 +09:00
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2008-11-12 14:34:12 -05:00
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2012-03-01 17:24:58 +01:00
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2023-11-22 12:17:47 -05:00
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2020-07-23 00:18:04 +09:00
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2012-03-01 17:24:58 +01:00
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2008-11-12 14:34:12 -05:00
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2011-04-12 10:43:21 +02:00
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2021-07-26 14:15:26 +02:00
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2011-04-12 10:43:21 +02:00
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2021-07-26 14:15:26 +02:00
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2011-04-12 10:43:21 +02:00
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2021-07-26 14:15:26 +02:00
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2011-04-12 10:43:21 +02:00
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btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
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2018-11-03 16:39:28 +01:00
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btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
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2025-10-24 12:21:42 +02:00
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|
btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
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2019-07-03 15:32:59 +03:00
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|
btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
|
|
|
|
2019-06-18 20:00:11 +02:00
|
|
|
|
btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
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2014-11-03 08:56:50 -05:00
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2016-05-19 21:18:45 -04:00
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2014-11-03 08:56:50 -05:00
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2011-11-28 16:43:00 +08:00
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2014-11-03 08:56:50 -05:00
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btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
|
|
|
|
2013-10-31 10:32:18 +05:30
|
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|
|
2017-06-15 15:04:04 +02:00
|
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|
|
btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
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|
|
2016-05-19 21:18:45 -04:00
|
|
|
|
2017-05-17 11:38:35 -04:00
|
|
|
|
2019-06-18 20:00:11 +02:00
|
|
|
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|
|
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|
|
2019-06-18 20:00:13 +02:00
|
|
|
|
2011-12-14 20:12:02 -05:00
|
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|
2022-06-23 16:57:00 +02:00
|
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|
2019-06-18 20:00:13 +02:00
|
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|
2011-12-14 20:12:02 -05:00
|
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|
btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
|
|
|
|
2019-06-18 20:00:11 +02:00
|
|
|
|
|
|
|
|
|
btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
|
|
|
|
2014-11-03 08:56:50 -05:00
|
|
|
|
|
|
|
|
|
2017-12-04 12:54:53 +08:00
|
|
|
|
|
|
|
|
|
2017-12-04 12:54:55 +08:00
|
|
|
|
|
|
|
|
|
btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
|
|
|
|
|
|
|
|
|
2014-11-03 08:56:50 -05:00
|
|
|
|
|
|
|
|
|
|
|
|
|
|
btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
|
|
|
|
|
|
|
|
|
|
|
|
|
|
2019-07-03 15:32:59 +03:00
|
|
|
|
btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
|
|
|
|
|
|
|
|
|
2022-06-13 15:06:34 +08:00
|
|
|
|
|
|
|
|
|
btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
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2022-06-13 15:06:34 +08:00
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|
btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
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2022-06-13 15:06:34 +08:00
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2019-07-03 15:32:59 +03:00
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btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
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2014-11-03 08:56:50 -05:00
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btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
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2019-07-03 15:32:59 +03:00
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2011-12-14 20:12:02 -05:00
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btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
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2011-12-14 20:12:02 -05:00
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|
btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
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2011-12-14 20:12:02 -05:00
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|
btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
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2014-07-01 16:21:33 +02:00
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2017-06-15 01:30:06 +02:00
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2014-07-01 16:21:33 +02:00
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2015-10-10 17:59:53 +02:00
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2014-07-01 16:21:33 +02:00
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2007-04-19 21:01:03 -04:00
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2011-11-17 15:40:49 -05:00
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Btrfs: make df be a little bit more understandable
The way we report df usage is way confusing for everybody, including some other
utilities (bacula for one). So this patch makes df a little bit more
understandable. First we make used actually count the total amount of used
space in all space info's. This will give us a real view of how much disk space
is in use. Second, for blocks available, only count data space. This makes
things like bacula work because it says 0 when you can no longer write anymore
data to the disk. I think this is a nice compromise, since you will end up with
something like the following
[root@alpha ~]# df -h
Filesystem Size Used Avail Use% Mounted on
/dev/mapper/VolGroup-lv_root
148G 30G 111G 21% /
/dev/sda1 194M 116M 68M 64% /boot
tmpfs 985M 12K 985M 1% /dev/shm
/dev/mapper/VolGroup-LogVol02
145G 140G 0 100% /mnt/btrfs-test
Compare this with btrfsctl -i output
[root@alpha btrfs-progs-unstable]# ./btrfsctl -i /mnt/btrfs-test/
Metadata, DUP: total=4.62GB, used=2.46GB
System, DUP: total=8.00MB, used=24.00KB
Data: total=134.80GB, used=134.80GB
Metadata: total=8.00MB, used=0.00
System: total=4.00MB, used=0.00
operation complete
This way we show that there is no more data space to be used, but we have
another 5GB of space left for metadata. Thanks,
Signed-off-by: Josef Bacik <josef@redhat.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2010-03-05 21:59:21 +00:00
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btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
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|
2015-10-10 17:59:53 +02:00
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2020-07-01 21:19:09 +02:00
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2018-10-30 16:43:24 +02:00
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2014-07-01 16:21:33 +02:00
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btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
|
|
|
|
2015-10-10 17:59:53 +02:00
|
|
|
|
2016-03-30 21:53:38 +01:00
|
|
|
|
2007-04-19 21:01:03 -04:00
|
|
|
|
2020-09-01 17:40:37 -04:00
|
|
|
|
btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
|
|
|
|
2014-07-01 16:21:33 +02:00
|
|
|
|
|
|
|
|
|
btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
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2014-07-01 16:21:33 +02:00
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2018-07-13 20:46:30 +02:00
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2014-07-01 16:21:33 +02:00
|
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|
|
btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
|
|
|
|
2016-03-30 21:53:38 +01:00
|
|
|
|
|
|
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|
|
2023-01-17 10:03:21 +00:00
|
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|
2016-03-30 21:53:38 +01:00
|
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btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
|
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2010-05-16 10:46:24 -04:00
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2010-10-14 14:52:27 -04:00
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2014-07-01 16:21:33 +02:00
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2016-03-30 23:18:14 +01:00
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2014-07-01 16:21:33 +02:00
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2014-11-14 15:05:06 +01:00
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2016-06-21 21:16:51 -04:00
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2014-11-03 08:56:50 -05:00
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|
btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
|
|
|
|
2014-07-01 16:21:33 +02:00
|
|
|
|
btrfs: fix wrong free space information of btrfs
When we store data by raid profile in btrfs with two or more different size
disks, df command shows there is some free space in the filesystem, but the
user can not write any data in fact, df command shows the wrong free space
information of btrfs.
# mkfs.btrfs -d raid1 /dev/sda9 /dev/sda10
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 28.00KB
devid 1 size 5.01GB used 2.03GB path /dev/sda9
devid 2 size 10.00GB used 2.01GB path /dev/sda10
# btrfs device scan /dev/sda9 /dev/sda10
# mount /dev/sda9 /mnt
# dd if=/dev/zero of=tmpfile0 bs=4K count=9999999999
(fill the filesystem)
# sync
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 5.4G 62% /mnt
# btrfs-show
Label: none uuid: a95cd49e-6e33-45b8-8741-a36153ce4b64
Total devices 2 FS bytes used 3.99GB
devid 1 size 5.01GB used 5.01GB path /dev/sda9
devid 2 size 10.00GB used 4.99GB path /dev/sda10
It is because btrfs cannot allocate chunks when one of the pairing disks has
no space, the free space on the other disks can not be used for ever, and should
be subtracted from the total space, but btrfs doesn't subtract this space from
the total. It is strange to the user.
This patch fixes it by calcing the free space that can be used to allocate
chunks.
Implementation:
1. get all the devices free space, and align them by stripe length.
2. sort the devices by the free space.
3. check the free space of the devices,
3.1. if it is not zero, and then check the number of the devices that has
more free space than this device,
if the number of the devices is beyond the min stripe number, the free
space can be used, and add into total free space.
if the number of the devices is below the min stripe number, we can not
use the free space, the check ends.
3.2. if the free space is zero, check the next devices, goto 3.1
This implementation is just likely fake chunk allocation.
After appling this patch, df can show correct space information:
# df -TH
Filesystem Type Size Used Avail Use% Mounted on
/dev/sda9 btrfs 17G 8.6G 0 100% /mnt
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Chris Mason <chris.mason@oracle.com>
2011-01-05 10:07:31 +00:00
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2009-01-05 21:25:51 -05:00
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2015-10-10 17:59:53 +02:00
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2017-10-16 16:48:40 +03:00
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2015-10-10 17:59:53 +02:00
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2020-01-31 09:31:05 -05:00
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2023-09-18 10:34:51 -04:00
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2015-10-10 17:59:53 +02:00
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2014-07-01 16:21:33 +02:00
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2024-01-16 17:33:20 +01:00
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2014-07-01 16:21:33 +02:00
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2008-08-18 12:01:52 +01:00
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2009-01-05 21:25:51 -05:00
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2008-08-18 12:01:52 +01:00
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2008-08-18 13:10:20 +01:00
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2024-04-15 16:16:23 -04:00
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2008-08-18 13:10:20 +01:00
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2007-04-19 21:01:03 -04:00
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2007-04-24 11:52:22 -04:00
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2023-11-22 12:17:48 -05:00
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btrfs: restrict writes to opened btrfs devices
[FLAG EXCLUSION]
Commit ead622674df5 ("btrfs: Do not restrict writes to btrfs devices")
removes the BLK_OPEN_RESTRICT_WRITES flag when opening the devices
during mount. This was an exception at the time as it depended on other
patches.
[REASON TO EXCLUDE THAT FLAG]
Btrfs needs to call btrfs_scan_one_device() to determine the fsid, no
matter if we're mounting a new fs or an existing one.
But if a fs is already mounted and the BLK_OPEN_RESTRICT_WRITES is
honored, meaning no other write open is allowed for the block device.
Then we want to mount a subvolume of the mounted fs to another mount
point, we will call btrfs_scan_one_device() again, but it will fail due
to the BLK_OPEN_RESTRICT_WRITES flag (no more write open allowed),
causing only one mount point for the fs.
Thus at that time, we had to exclude the BLK_OPEN_RESTRICT_WRITES to
allow multiple mount points for one fs.
[WHY IT'S SAFE NOW]
The root problem is, we do not need to nor should use BLK_OPEN_WRITE for
btrfs_scan_one_device().
That function is only to read out the super block, no write at all, and
BLK_OPEN_WRITE is only going to cause problems for such usage.
The root problem has been fixed by patch "btrfs: always open the device
read-only in btrfs_scan_one_device", so btrfs_scan_one_device() will
always work no matter if the device is opened with
BLK_OPEN_RESTRICT_WRITES.
[ENHANCEMENT]
Just remove the btrfs_open_mode(), as the only call site can be replaced
with regular sb_open_mode().
Reviewed-by: Christian Brauner <brauner@kernel.org>
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2025-07-04 19:08:03 +09:30
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2023-11-22 12:17:48 -05:00
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2025-06-11 12:02:59 +02:00
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2023-11-22 12:17:48 -05:00
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btrfs: delay btrfs_open_devices() until super block is created
Currently we always call btrfs_open_devices() before creating the
super block.
It's fine for now because:
- No blk_holder_ops is provided
- btrfs_fs_type is used as a holder
This means no matter who wins the device opening race, the holder will be
the same thus not affecting the later sget_fc() race.
And since no blk_holder_ops is provided, no bdev operation is depending on
the holder.
But this will no longer be true if we want to implement a proper
blk_holder_ops using fs_holder_ops.
This means we will need a proper super block as the bdev holder.
To prepare for such change:
- Add btrfs_fs_devices::holding member
This will prevent btrfs_free_stale_devices() and btrfs_close_device()
from deleting the fs_devices when there is another process trying to
mount the fs.
Along with the new member, here come the two helpers,
btrfs_fs_devices_inc_holding() and btrfs_fs_devices_dec_holding().
This will allow us to hold fs_devices without opening it.
This is needed because we cannot hold uuid_mutex while calling
sget_fc(), this will reverse the lock sequence with s_umount, causing
a lockdep warning.
- Delay btrfs_open_devices() until a super block is returned
This means we have to hold the initial fs_devices first, then unlock
uuid_mutex, call sget_fc(), then re-lock uuid_mutex, and decrease the
holding number.
For new super block case, we continue to btrfs_open_devices() with
uuid_mutex hold.
For existing super block case, we can unlock uuid_mutex and continue.
Although this means a more complex error handling path, as if we
didn't call btrfs_open_devices() (either got an existing sb, or
sget_fc() failed), we cannot let btrfs_put_fs_info() cleanup the
fs_devices, as it can be freed at any time after we decrease the hold
on fs_devices and unlock uuid_mutex.
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2025-06-17 07:41:14 +09:30
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2023-11-22 12:17:48 -05:00
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btrfs: delay btrfs_open_devices() until super block is created
Currently we always call btrfs_open_devices() before creating the
super block.
It's fine for now because:
- No blk_holder_ops is provided
- btrfs_fs_type is used as a holder
This means no matter who wins the device opening race, the holder will be
the same thus not affecting the later sget_fc() race.
And since no blk_holder_ops is provided, no bdev operation is depending on
the holder.
But this will no longer be true if we want to implement a proper
blk_holder_ops using fs_holder_ops.
This means we will need a proper super block as the bdev holder.
To prepare for such change:
- Add btrfs_fs_devices::holding member
This will prevent btrfs_free_stale_devices() and btrfs_close_device()
from deleting the fs_devices when there is another process trying to
mount the fs.
Along with the new member, here come the two helpers,
btrfs_fs_devices_inc_holding() and btrfs_fs_devices_dec_holding().
This will allow us to hold fs_devices without opening it.
This is needed because we cannot hold uuid_mutex while calling
sget_fc(), this will reverse the lock sequence with s_umount, causing
a lockdep warning.
- Delay btrfs_open_devices() until a super block is returned
This means we have to hold the initial fs_devices first, then unlock
uuid_mutex, call sget_fc(), then re-lock uuid_mutex, and decrease the
holding number.
For new super block case, we continue to btrfs_open_devices() with
uuid_mutex hold.
For existing super block case, we can unlock uuid_mutex and continue.
Although this means a more complex error handling path, as if we
didn't call btrfs_open_devices() (either got an existing sb, or
sget_fc() failed), we cannot let btrfs_put_fs_info() cleanup the
fs_devices, as it can be freed at any time after we decrease the hold
on fs_devices and unlock uuid_mutex.
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2025-06-17 07:41:14 +09:30
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2025-06-11 12:03:00 +02:00
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btrfs: delay btrfs_open_devices() until super block is created
Currently we always call btrfs_open_devices() before creating the
super block.
It's fine for now because:
- No blk_holder_ops is provided
- btrfs_fs_type is used as a holder
This means no matter who wins the device opening race, the holder will be
the same thus not affecting the later sget_fc() race.
And since no blk_holder_ops is provided, no bdev operation is depending on
the holder.
But this will no longer be true if we want to implement a proper
blk_holder_ops using fs_holder_ops.
This means we will need a proper super block as the bdev holder.
To prepare for such change:
- Add btrfs_fs_devices::holding member
This will prevent btrfs_free_stale_devices() and btrfs_close_device()
from deleting the fs_devices when there is another process trying to
mount the fs.
Along with the new member, here come the two helpers,
btrfs_fs_devices_inc_holding() and btrfs_fs_devices_dec_holding().
This will allow us to hold fs_devices without opening it.
This is needed because we cannot hold uuid_mutex while calling
sget_fc(), this will reverse the lock sequence with s_umount, causing
a lockdep warning.
- Delay btrfs_open_devices() until a super block is returned
This means we have to hold the initial fs_devices first, then unlock
uuid_mutex, call sget_fc(), then re-lock uuid_mutex, and decrease the
holding number.
For new super block case, we continue to btrfs_open_devices() with
uuid_mutex hold.
For existing super block case, we can unlock uuid_mutex and continue.
Although this means a more complex error handling path, as if we
didn't call btrfs_open_devices() (either got an existing sb, or
sget_fc() failed), we cannot let btrfs_put_fs_info() cleanup the
fs_devices, as it can be freed at any time after we decrease the hold
on fs_devices and unlock uuid_mutex.
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2025-06-17 07:41:14 +09:30
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2023-11-22 12:17:48 -05:00
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2025-06-16 08:14:25 +09:30
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2025-06-11 12:03:00 +02:00
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2025-06-16 08:14:25 +09:30
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2025-06-11 12:03:00 +02:00
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btrfs: delay btrfs_open_devices() until super block is created
Currently we always call btrfs_open_devices() before creating the
super block.
It's fine for now because:
- No blk_holder_ops is provided
- btrfs_fs_type is used as a holder
This means no matter who wins the device opening race, the holder will be
the same thus not affecting the later sget_fc() race.
And since no blk_holder_ops is provided, no bdev operation is depending on
the holder.
But this will no longer be true if we want to implement a proper
blk_holder_ops using fs_holder_ops.
This means we will need a proper super block as the bdev holder.
To prepare for such change:
- Add btrfs_fs_devices::holding member
This will prevent btrfs_free_stale_devices() and btrfs_close_device()
from deleting the fs_devices when there is another process trying to
mount the fs.
Along with the new member, here come the two helpers,
btrfs_fs_devices_inc_holding() and btrfs_fs_devices_dec_holding().
This will allow us to hold fs_devices without opening it.
This is needed because we cannot hold uuid_mutex while calling
sget_fc(), this will reverse the lock sequence with s_umount, causing
a lockdep warning.
- Delay btrfs_open_devices() until a super block is returned
This means we have to hold the initial fs_devices first, then unlock
uuid_mutex, call sget_fc(), then re-lock uuid_mutex, and decrease the
holding number.
For new super block case, we continue to btrfs_open_devices() with
uuid_mutex hold.
For existing super block case, we can unlock uuid_mutex and continue.
Although this means a more complex error handling path, as if we
didn't call btrfs_open_devices() (either got an existing sb, or
sget_fc() failed), we cannot let btrfs_put_fs_info() cleanup the
fs_devices, as it can be freed at any time after we decrease the hold
on fs_devices and unlock uuid_mutex.
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2025-06-17 07:41:14 +09:30
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btrfs: fix mount failure due to remount races
[BUG]
The following reproducer can cause btrfs mount to fail:
dev="/dev/test/scratch1"
mnt1="/mnt/test"
mnt2="/mnt/scratch"
mkfs.btrfs -f $dev
mount $dev $mnt1
btrfs subvolume create $mnt1/subvol1
btrfs subvolume create $mnt1/subvol2
umount $mnt1
mount $dev $mnt1 -o subvol=subvol1
while mount -o remount,ro $mnt1; do mount -o remount,rw $mnt1; done &
bg=$!
while mount $dev $mnt2 -o subvol=subvol2; do umount $mnt2; done
kill $bg
wait
umount -R $mnt1
umount -R $mnt2
The script will fail with the following error:
mount: /mnt/scratch: /dev/mapper/test-scratch1 already mounted on /mnt/test.
dmesg(1) may have more information after failed mount system call.
umount: /mnt/test: target is busy.
umount: /mnt/scratch/: not mounted
And there is no kernel error message.
[CAUSE]
During the btrfs mount, to support mounting different subvolumes with
different RO/RW flags, we need to detect that and retry if needed:
Retry with matching RO flags if the initial mount fail with -EBUSY.
The problem is, during that retry we do not hold any super block lock
(s_umount), this means there can be a remount process changing the RO
flags of the original fs super block.
If so, we can have an EBUSY error during retry. And this time we treat
any failure as an error, without any retry and cause the above EBUSY
mount failure.
[FIX]
The current retry behavior is racy because we do not have a super block
thus no way to hold s_umount to prevent the race with remount.
Solve the root problem by allowing fc->sb_flags to mismatch from the
sb->s_flags at btrfs_get_tree_super().
Then at the re-entry point btrfs_get_tree_subvol(), manually check the
fc->s_flags against sb->s_flags, if it's a RO->RW mismatch, then
reconfigure with s_umount lock hold.
Reported-by: Enno Gotthold <egotthold@suse.com>
Reported-by: Fabian Vogt <fvogt@suse.com>
[ Special thanks for the reproducer and early analysis pointing to btrfs. ]
Fixes: f044b318675f ("btrfs: handle the ro->rw transition for mounting different subvolumes")
Link: https://bugzilla.suse.com/show_bug.cgi?id=1231836
Signed-off-by: Qu Wenruo <wqu@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2024-10-30 11:25:48 +10:30
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2023-11-22 12:17:48 -05:00
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btrfs: delay btrfs_open_devices() until super block is created
Currently we always call btrfs_open_devices() before creating the
super block.
It's fine for now because:
- No blk_holder_ops is provided
- btrfs_fs_type is used as a holder
This means no matter who wins the device opening race, the holder will be
the same thus not affecting the later sget_fc() race.
And since no blk_holder_ops is provided, no bdev operation is depending on
the holder.
But this will no longer be true if we want to implement a proper
blk_holder_ops using fs_holder_ops.
This means we will need a proper super block as the bdev holder.
To prepare for such change:
- Add btrfs_fs_devices::holding member
This will prevent btrfs_free_stale_devices() and btrfs_close_device()
from deleting the fs_devices when there is another process trying to
mount the fs.
Along with the new member, here come the two helpers,
btrfs_fs_devices_inc_holding() and btrfs_fs_devices_dec_holding().
This will allow us to hold fs_devices without opening it.
This is needed because we cannot hold uuid_mutex while calling
sget_fc(), this will reverse the lock sequence with s_umount, causing
a lockdep warning.
- Delay btrfs_open_devices() until a super block is returned
This means we have to hold the initial fs_devices first, then unlock
uuid_mutex, call sget_fc(), then re-lock uuid_mutex, and decrease the
holding number.
For new super block case, we continue to btrfs_open_devices() with
uuid_mutex hold.
For existing super block case, we can unlock uuid_mutex and continue.
Although this means a more complex error handling path, as if we
didn't call btrfs_open_devices() (either got an existing sb, or
sget_fc() failed), we cannot let btrfs_put_fs_info() cleanup the
fs_devices, as it can be freed at any time after we decrease the hold
on fs_devices and unlock uuid_mutex.
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2025-06-17 07:41:14 +09:30
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2025-06-16 08:14:25 +09:30
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btrfs: delay btrfs_open_devices() until super block is created
Currently we always call btrfs_open_devices() before creating the
super block.
It's fine for now because:
- No blk_holder_ops is provided
- btrfs_fs_type is used as a holder
This means no matter who wins the device opening race, the holder will be
the same thus not affecting the later sget_fc() race.
And since no blk_holder_ops is provided, no bdev operation is depending on
the holder.
But this will no longer be true if we want to implement a proper
blk_holder_ops using fs_holder_ops.
This means we will need a proper super block as the bdev holder.
To prepare for such change:
- Add btrfs_fs_devices::holding member
This will prevent btrfs_free_stale_devices() and btrfs_close_device()
from deleting the fs_devices when there is another process trying to
mount the fs.
Along with the new member, here come the two helpers,
btrfs_fs_devices_inc_holding() and btrfs_fs_devices_dec_holding().
This will allow us to hold fs_devices without opening it.
This is needed because we cannot hold uuid_mutex while calling
sget_fc(), this will reverse the lock sequence with s_umount, causing
a lockdep warning.
- Delay btrfs_open_devices() until a super block is returned
This means we have to hold the initial fs_devices first, then unlock
uuid_mutex, call sget_fc(), then re-lock uuid_mutex, and decrease the
holding number.
For new super block case, we continue to btrfs_open_devices() with
uuid_mutex hold.
For existing super block case, we can unlock uuid_mutex and continue.
Although this means a more complex error handling path, as if we
didn't call btrfs_open_devices() (either got an existing sb, or
sget_fc() failed), we cannot let btrfs_put_fs_info() cleanup the
fs_devices, as it can be freed at any time after we decrease the hold
on fs_devices and unlock uuid_mutex.
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2025-06-17 07:41:14 +09:30
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2025-06-11 12:03:03 +02:00
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btrfs: delay btrfs_open_devices() until super block is created
Currently we always call btrfs_open_devices() before creating the
super block.
It's fine for now because:
- No blk_holder_ops is provided
- btrfs_fs_type is used as a holder
This means no matter who wins the device opening race, the holder will be
the same thus not affecting the later sget_fc() race.
And since no blk_holder_ops is provided, no bdev operation is depending on
the holder.
But this will no longer be true if we want to implement a proper
blk_holder_ops using fs_holder_ops.
This means we will need a proper super block as the bdev holder.
To prepare for such change:
- Add btrfs_fs_devices::holding member
This will prevent btrfs_free_stale_devices() and btrfs_close_device()
from deleting the fs_devices when there is another process trying to
mount the fs.
Along with the new member, here come the two helpers,
btrfs_fs_devices_inc_holding() and btrfs_fs_devices_dec_holding().
This will allow us to hold fs_devices without opening it.
This is needed because we cannot hold uuid_mutex while calling
sget_fc(), this will reverse the lock sequence with s_umount, causing
a lockdep warning.
- Delay btrfs_open_devices() until a super block is returned
This means we have to hold the initial fs_devices first, then unlock
uuid_mutex, call sget_fc(), then re-lock uuid_mutex, and decrease the
holding number.
For new super block case, we continue to btrfs_open_devices() with
uuid_mutex hold.
For existing super block case, we can unlock uuid_mutex and continue.
Although this means a more complex error handling path, as if we
didn't call btrfs_open_devices() (either got an existing sb, or
sget_fc() failed), we cannot let btrfs_put_fs_info() cleanup the
fs_devices, as it can be freed at any time after we decrease the hold
on fs_devices and unlock uuid_mutex.
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2025-06-17 07:41:14 +09:30
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btrfs: only set the device specific options after devices are opened
[BUG]
With v6.17-rc kernels, btrfs will always set 'ssd' mount option even if
the block device is not a rotating one:
# cat /sys/block/sdd/queue/rotational
1
# cat /etc/fstab:
LABEL=DATA2 /data2 btrfs rw,relatime,space_cache=v2,subvolid=5,subvol=/,nofail,nosuid,nodev 0 0
# mount
[...]
/dev/sdd on /data2 type btrfs (rw,nosuid,nodev,relatime,ssd,space_cache=v2,subvolid=5,subvol=/)
[CAUSE]
The 'ssd' mount option is set by set_device_specific_options(), and it
expects that if there is any rotating device in the btrfs, it will set
fs_devices::rotating.
However after commit bddf57a70781 ("btrfs: delay btrfs_open_devices()
until super block is created"), the device opening is delayed until the
super block is created.
But the timing of set_device_specific_options() is still left as is,
this makes the function be called without any device opened.
Since no device is opened, thus fs_devices::rotating will never be set,
making btrfs incorrectly set 'ssd' mount option.
[FIX]
Only call set_device_specific_options() after btrfs_open_devices().
Also only call set_device_specific_options() after a new mount, if we're
mounting a mounted btrfs, there is no need to set the device specific
mount options again.
Reported-by: HAN Yuwei <hrx@bupt.moe>
Link: https://lore.kernel.org/linux-btrfs/C8FF75669DFFC3C5+5f93bf8a-80a0-48a6-81bf-4ec890abc99a@bupt.moe/
Fixes: bddf57a70781 ("btrfs: delay btrfs_open_devices() until super block is created")
CC: stable@vger.kernel.org # 6.17
Signed-off-by: Qu Wenruo <wqu@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2025-09-26 14:20:11 +09:30
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btrfs: delay btrfs_open_devices() until super block is created
Currently we always call btrfs_open_devices() before creating the
super block.
It's fine for now because:
- No blk_holder_ops is provided
- btrfs_fs_type is used as a holder
This means no matter who wins the device opening race, the holder will be
the same thus not affecting the later sget_fc() race.
And since no blk_holder_ops is provided, no bdev operation is depending on
the holder.
But this will no longer be true if we want to implement a proper
blk_holder_ops using fs_holder_ops.
This means we will need a proper super block as the bdev holder.
To prepare for such change:
- Add btrfs_fs_devices::holding member
This will prevent btrfs_free_stale_devices() and btrfs_close_device()
from deleting the fs_devices when there is another process trying to
mount the fs.
Along with the new member, here come the two helpers,
btrfs_fs_devices_inc_holding() and btrfs_fs_devices_dec_holding().
This will allow us to hold fs_devices without opening it.
This is needed because we cannot hold uuid_mutex while calling
sget_fc(), this will reverse the lock sequence with s_umount, causing
a lockdep warning.
- Delay btrfs_open_devices() until a super block is returned
This means we have to hold the initial fs_devices first, then unlock
uuid_mutex, call sget_fc(), then re-lock uuid_mutex, and decrease the
holding number.
For new super block case, we continue to btrfs_open_devices() with
uuid_mutex hold.
For existing super block case, we can unlock uuid_mutex and continue.
Although this means a more complex error handling path, as if we
didn't call btrfs_open_devices() (either got an existing sb, or
sget_fc() failed), we cannot let btrfs_put_fs_info() cleanup the
fs_devices, as it can be freed at any time after we decrease the hold
on fs_devices and unlock uuid_mutex.
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2025-06-17 07:41:14 +09:30
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2023-11-22 12:17:48 -05:00
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2024-10-09 16:32:17 +02:00
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btrfs: fix mount failure due to remount races
[BUG]
The following reproducer can cause btrfs mount to fail:
dev="/dev/test/scratch1"
mnt1="/mnt/test"
mnt2="/mnt/scratch"
mkfs.btrfs -f $dev
mount $dev $mnt1
btrfs subvolume create $mnt1/subvol1
btrfs subvolume create $mnt1/subvol2
umount $mnt1
mount $dev $mnt1 -o subvol=subvol1
while mount -o remount,ro $mnt1; do mount -o remount,rw $mnt1; done &
bg=$!
while mount $dev $mnt2 -o subvol=subvol2; do umount $mnt2; done
kill $bg
wait
umount -R $mnt1
umount -R $mnt2
The script will fail with the following error:
mount: /mnt/scratch: /dev/mapper/test-scratch1 already mounted on /mnt/test.
dmesg(1) may have more information after failed mount system call.
umount: /mnt/test: target is busy.
umount: /mnt/scratch/: not mounted
And there is no kernel error message.
[CAUSE]
During the btrfs mount, to support mounting different subvolumes with
different RO/RW flags, we need to detect that and retry if needed:
Retry with matching RO flags if the initial mount fail with -EBUSY.
The problem is, during that retry we do not hold any super block lock
(s_umount), this means there can be a remount process changing the RO
flags of the original fs super block.
If so, we can have an EBUSY error during retry. And this time we treat
any failure as an error, without any retry and cause the above EBUSY
mount failure.
[FIX]
The current retry behavior is racy because we do not have a super block
thus no way to hold s_umount to prevent the race with remount.
Solve the root problem by allowing fc->sb_flags to mismatch from the
sb->s_flags at btrfs_get_tree_super().
Then at the re-entry point btrfs_get_tree_subvol(), manually check the
fc->s_flags against sb->s_flags, if it's a RO->RW mismatch, then
reconfigure with s_umount lock hold.
Reported-by: Enno Gotthold <egotthold@suse.com>
Reported-by: Fabian Vogt <fvogt@suse.com>
[ Special thanks for the reproducer and early analysis pointing to btrfs. ]
Fixes: f044b318675f ("btrfs: handle the ro->rw transition for mounting different subvolumes")
Link: https://bugzilla.suse.com/show_bug.cgi?id=1231836
Signed-off-by: Qu Wenruo <wqu@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2024-10-30 11:25:48 +10:30
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2023-11-22 12:17:48 -05:00
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2023-11-22 12:17:53 -05:00
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2023-11-22 12:17:48 -05:00
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2023-11-22 12:17:49 -05:00
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btrfs: fix per-subvolume RO/RW flags with new mount API
[BUG]
With util-linux 2.40.2, the 'mount' utility is already utilizing the new
mount API. e.g:
# strace mount -o subvol=subv1,ro /dev/test/scratch1 /mnt/test/
...
fsconfig(3, FSCONFIG_SET_STRING, "source", "/dev/mapper/test-scratch1", 0) = 0
fsconfig(3, FSCONFIG_SET_STRING, "subvol", "subv1", 0) = 0
fsconfig(3, FSCONFIG_SET_FLAG, "ro", NULL, 0) = 0
fsconfig(3, FSCONFIG_CMD_CREATE, NULL, NULL, 0) = 0
fsmount(3, FSMOUNT_CLOEXEC, 0) = 4
mount_setattr(4, "", AT_EMPTY_PATH, {attr_set=MOUNT_ATTR_RDONLY, attr_clr=0, propagation=0 /* MS_??? */, userns_fd=0}, 32) = 0
move_mount(4, "", AT_FDCWD, "/mnt/test", MOVE_MOUNT_F_EMPTY_PATH) = 0
But this leads to a new problem, that per-subvolume RO/RW mount no
longer works, if the initial mount is RO:
# mount -o subvol=subv1,ro /dev/test/scratch1 /mnt/test
# mount -o rw,subvol=subv2 /dev/test/scratch1 /mnt/scratch
# mount | grep mnt
/dev/mapper/test-scratch1 on /mnt/test type btrfs (ro,relatime,discard=async,space_cache=v2,subvolid=256,subvol=/subv1)
/dev/mapper/test-scratch1 on /mnt/scratch type btrfs (ro,relatime,discard=async,space_cache=v2,subvolid=257,subvol=/subv2)
# touch /mnt/scratch/foobar
touch: cannot touch '/mnt/scratch/foobar': Read-only file system
This is a common use cases on distros.
[CAUSE]
We have a workaround for remount to handle the RO->RW change, but if the
mount is using the new mount API, we do not do that, and rely on the
mount tool NOT to set the ro flag.
But that's not how the mount tool is doing for the new API:
fsconfig(3, FSCONFIG_SET_STRING, "source", "/dev/mapper/test-scratch1", 0) = 0
fsconfig(3, FSCONFIG_SET_STRING, "subvol", "subv1", 0) = 0
fsconfig(3, FSCONFIG_SET_FLAG, "ro", NULL, 0) = 0 <<<< Setting RO flag for super block
fsconfig(3, FSCONFIG_CMD_CREATE, NULL, NULL, 0) = 0
fsmount(3, FSMOUNT_CLOEXEC, 0) = 4
mount_setattr(4, "", AT_EMPTY_PATH, {attr_set=MOUNT_ATTR_RDONLY, attr_clr=0, propagation=0 /* MS_??? */, userns_fd=0}, 32) = 0
move_mount(4, "", AT_FDCWD, "/mnt/test", MOVE_MOUNT_F_EMPTY_PATH) = 0
This means we will set the super block RO at the first mount.
Later RW mount will not try to reconfigure the fs to RW because the
mount tool is already using the new API.
This totally breaks the per-subvolume RO/RW mount behavior.
[FIX]
Do not skip the reconfiguration even if using the new API. The old
comments are just expecting any mount tool to properly skip the RO flag
set even if we specify "ro", which is not the reality.
Update the comments regarding the backward compatibility on the kernel
level so it works with old and new mount utilities.
CC: stable@vger.kernel.org # 6.8+
Fixes: f044b318675f ("btrfs: handle the ro->rw transition for mounting different subvolumes")
Signed-off-by: Qu Wenruo <wqu@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2024-10-30 11:25:47 +10:30
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2023-11-22 12:17:49 -05:00
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2025-05-06 20:58:26 +01:00
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2023-11-22 12:17:49 -05:00
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btrfs: fix mount failure due to remount races
[BUG]
The following reproducer can cause btrfs mount to fail:
dev="/dev/test/scratch1"
mnt1="/mnt/test"
mnt2="/mnt/scratch"
mkfs.btrfs -f $dev
mount $dev $mnt1
btrfs subvolume create $mnt1/subvol1
btrfs subvolume create $mnt1/subvol2
umount $mnt1
mount $dev $mnt1 -o subvol=subvol1
while mount -o remount,ro $mnt1; do mount -o remount,rw $mnt1; done &
bg=$!
while mount $dev $mnt2 -o subvol=subvol2; do umount $mnt2; done
kill $bg
wait
umount -R $mnt1
umount -R $mnt2
The script will fail with the following error:
mount: /mnt/scratch: /dev/mapper/test-scratch1 already mounted on /mnt/test.
dmesg(1) may have more information after failed mount system call.
umount: /mnt/test: target is busy.
umount: /mnt/scratch/: not mounted
And there is no kernel error message.
[CAUSE]
During the btrfs mount, to support mounting different subvolumes with
different RO/RW flags, we need to detect that and retry if needed:
Retry with matching RO flags if the initial mount fail with -EBUSY.
The problem is, during that retry we do not hold any super block lock
(s_umount), this means there can be a remount process changing the RO
flags of the original fs super block.
If so, we can have an EBUSY error during retry. And this time we treat
any failure as an error, without any retry and cause the above EBUSY
mount failure.
[FIX]
The current retry behavior is racy because we do not have a super block
thus no way to hold s_umount to prevent the race with remount.
Solve the root problem by allowing fc->sb_flags to mismatch from the
sb->s_flags at btrfs_get_tree_super().
Then at the re-entry point btrfs_get_tree_subvol(), manually check the
fc->s_flags against sb->s_flags, if it's a RO->RW mismatch, then
reconfigure with s_umount lock hold.
Reported-by: Enno Gotthold <egotthold@suse.com>
Reported-by: Fabian Vogt <fvogt@suse.com>
[ Special thanks for the reproducer and early analysis pointing to btrfs. ]
Fixes: f044b318675f ("btrfs: handle the ro->rw transition for mounting different subvolumes")
Link: https://bugzilla.suse.com/show_bug.cgi?id=1231836
Signed-off-by: Qu Wenruo <wqu@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2024-10-30 11:25:48 +10:30
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2023-11-22 12:17:49 -05:00
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2025-05-06 20:58:26 +01:00
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btrfs: fix mount failure due to remount races
[BUG]
The following reproducer can cause btrfs mount to fail:
dev="/dev/test/scratch1"
mnt1="/mnt/test"
mnt2="/mnt/scratch"
mkfs.btrfs -f $dev
mount $dev $mnt1
btrfs subvolume create $mnt1/subvol1
btrfs subvolume create $mnt1/subvol2
umount $mnt1
mount $dev $mnt1 -o subvol=subvol1
while mount -o remount,ro $mnt1; do mount -o remount,rw $mnt1; done &
bg=$!
while mount $dev $mnt2 -o subvol=subvol2; do umount $mnt2; done
kill $bg
wait
umount -R $mnt1
umount -R $mnt2
The script will fail with the following error:
mount: /mnt/scratch: /dev/mapper/test-scratch1 already mounted on /mnt/test.
dmesg(1) may have more information after failed mount system call.
umount: /mnt/test: target is busy.
umount: /mnt/scratch/: not mounted
And there is no kernel error message.
[CAUSE]
During the btrfs mount, to support mounting different subvolumes with
different RO/RW flags, we need to detect that and retry if needed:
Retry with matching RO flags if the initial mount fail with -EBUSY.
The problem is, during that retry we do not hold any super block lock
(s_umount), this means there can be a remount process changing the RO
flags of the original fs super block.
If so, we can have an EBUSY error during retry. And this time we treat
any failure as an error, without any retry and cause the above EBUSY
mount failure.
[FIX]
The current retry behavior is racy because we do not have a super block
thus no way to hold s_umount to prevent the race with remount.
Solve the root problem by allowing fc->sb_flags to mismatch from the
sb->s_flags at btrfs_get_tree_super().
Then at the re-entry point btrfs_get_tree_subvol(), manually check the
fc->s_flags against sb->s_flags, if it's a RO->RW mismatch, then
reconfigure with s_umount lock hold.
Reported-by: Enno Gotthold <egotthold@suse.com>
Reported-by: Fabian Vogt <fvogt@suse.com>
[ Special thanks for the reproducer and early analysis pointing to btrfs. ]
Fixes: f044b318675f ("btrfs: handle the ro->rw transition for mounting different subvolumes")
Link: https://bugzilla.suse.com/show_bug.cgi?id=1231836
Signed-off-by: Qu Wenruo <wqu@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2024-10-30 11:25:48 +10:30
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2025-05-06 20:58:26 +01:00
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btrfs: fix mount failure due to remount races
[BUG]
The following reproducer can cause btrfs mount to fail:
dev="/dev/test/scratch1"
mnt1="/mnt/test"
mnt2="/mnt/scratch"
mkfs.btrfs -f $dev
mount $dev $mnt1
btrfs subvolume create $mnt1/subvol1
btrfs subvolume create $mnt1/subvol2
umount $mnt1
mount $dev $mnt1 -o subvol=subvol1
while mount -o remount,ro $mnt1; do mount -o remount,rw $mnt1; done &
bg=$!
while mount $dev $mnt2 -o subvol=subvol2; do umount $mnt2; done
kill $bg
wait
umount -R $mnt1
umount -R $mnt2
The script will fail with the following error:
mount: /mnt/scratch: /dev/mapper/test-scratch1 already mounted on /mnt/test.
dmesg(1) may have more information after failed mount system call.
umount: /mnt/test: target is busy.
umount: /mnt/scratch/: not mounted
And there is no kernel error message.
[CAUSE]
During the btrfs mount, to support mounting different subvolumes with
different RO/RW flags, we need to detect that and retry if needed:
Retry with matching RO flags if the initial mount fail with -EBUSY.
The problem is, during that retry we do not hold any super block lock
(s_umount), this means there can be a remount process changing the RO
flags of the original fs super block.
If so, we can have an EBUSY error during retry. And this time we treat
any failure as an error, without any retry and cause the above EBUSY
mount failure.
[FIX]
The current retry behavior is racy because we do not have a super block
thus no way to hold s_umount to prevent the race with remount.
Solve the root problem by allowing fc->sb_flags to mismatch from the
sb->s_flags at btrfs_get_tree_super().
Then at the re-entry point btrfs_get_tree_subvol(), manually check the
fc->s_flags against sb->s_flags, if it's a RO->RW mismatch, then
reconfigure with s_umount lock hold.
Reported-by: Enno Gotthold <egotthold@suse.com>
Reported-by: Fabian Vogt <fvogt@suse.com>
[ Special thanks for the reproducer and early analysis pointing to btrfs. ]
Fixes: f044b318675f ("btrfs: handle the ro->rw transition for mounting different subvolumes")
Link: https://bugzilla.suse.com/show_bug.cgi?id=1231836
Signed-off-by: Qu Wenruo <wqu@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2024-10-30 11:25:48 +10:30
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2023-11-22 12:17:49 -05:00
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2023-11-22 12:17:48 -05:00
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btrfs: fix mount failure due to remount races
[BUG]
The following reproducer can cause btrfs mount to fail:
dev="/dev/test/scratch1"
mnt1="/mnt/test"
mnt2="/mnt/scratch"
mkfs.btrfs -f $dev
mount $dev $mnt1
btrfs subvolume create $mnt1/subvol1
btrfs subvolume create $mnt1/subvol2
umount $mnt1
mount $dev $mnt1 -o subvol=subvol1
while mount -o remount,ro $mnt1; do mount -o remount,rw $mnt1; done &
bg=$!
while mount $dev $mnt2 -o subvol=subvol2; do umount $mnt2; done
kill $bg
wait
umount -R $mnt1
umount -R $mnt2
The script will fail with the following error:
mount: /mnt/scratch: /dev/mapper/test-scratch1 already mounted on /mnt/test.
dmesg(1) may have more information after failed mount system call.
umount: /mnt/test: target is busy.
umount: /mnt/scratch/: not mounted
And there is no kernel error message.
[CAUSE]
During the btrfs mount, to support mounting different subvolumes with
different RO/RW flags, we need to detect that and retry if needed:
Retry with matching RO flags if the initial mount fail with -EBUSY.
The problem is, during that retry we do not hold any super block lock
(s_umount), this means there can be a remount process changing the RO
flags of the original fs super block.
If so, we can have an EBUSY error during retry. And this time we treat
any failure as an error, without any retry and cause the above EBUSY
mount failure.
[FIX]
The current retry behavior is racy because we do not have a super block
thus no way to hold s_umount to prevent the race with remount.
Solve the root problem by allowing fc->sb_flags to mismatch from the
sb->s_flags at btrfs_get_tree_super().
Then at the re-entry point btrfs_get_tree_subvol(), manually check the
fc->s_flags against sb->s_flags, if it's a RO->RW mismatch, then
reconfigure with s_umount lock hold.
Reported-by: Enno Gotthold <egotthold@suse.com>
Reported-by: Fabian Vogt <fvogt@suse.com>
[ Special thanks for the reproducer and early analysis pointing to btrfs. ]
Fixes: f044b318675f ("btrfs: handle the ro->rw transition for mounting different subvolumes")
Link: https://bugzilla.suse.com/show_bug.cgi?id=1231836
Signed-off-by: Qu Wenruo <wqu@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2024-10-30 11:25:48 +10:30
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2023-11-22 12:17:48 -05:00
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2026-02-21 16:37:42 -08:00
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2023-11-22 12:17:48 -05:00
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2025-10-16 14:10:11 +08:00
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2023-11-22 12:17:48 -05:00
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2025-06-12 15:14:35 +09:30
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2025-05-06 20:58:26 +01:00
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btrfs: fix mount failure due to remount races
[BUG]
The following reproducer can cause btrfs mount to fail:
dev="/dev/test/scratch1"
mnt1="/mnt/test"
mnt2="/mnt/scratch"
mkfs.btrfs -f $dev
mount $dev $mnt1
btrfs subvolume create $mnt1/subvol1
btrfs subvolume create $mnt1/subvol2
umount $mnt1
mount $dev $mnt1 -o subvol=subvol1
while mount -o remount,ro $mnt1; do mount -o remount,rw $mnt1; done &
bg=$!
while mount $dev $mnt2 -o subvol=subvol2; do umount $mnt2; done
kill $bg
wait
umount -R $mnt1
umount -R $mnt2
The script will fail with the following error:
mount: /mnt/scratch: /dev/mapper/test-scratch1 already mounted on /mnt/test.
dmesg(1) may have more information after failed mount system call.
umount: /mnt/test: target is busy.
umount: /mnt/scratch/: not mounted
And there is no kernel error message.
[CAUSE]
During the btrfs mount, to support mounting different subvolumes with
different RO/RW flags, we need to detect that and retry if needed:
Retry with matching RO flags if the initial mount fail with -EBUSY.
The problem is, during that retry we do not hold any super block lock
(s_umount), this means there can be a remount process changing the RO
flags of the original fs super block.
If so, we can have an EBUSY error during retry. And this time we treat
any failure as an error, without any retry and cause the above EBUSY
mount failure.
[FIX]
The current retry behavior is racy because we do not have a super block
thus no way to hold s_umount to prevent the race with remount.
Solve the root problem by allowing fc->sb_flags to mismatch from the
sb->s_flags at btrfs_get_tree_super().
Then at the re-entry point btrfs_get_tree_subvol(), manually check the
fc->s_flags against sb->s_flags, if it's a RO->RW mismatch, then
reconfigure with s_umount lock hold.
Reported-by: Enno Gotthold <egotthold@suse.com>
Reported-by: Fabian Vogt <fvogt@suse.com>
[ Special thanks for the reproducer and early analysis pointing to btrfs. ]
Fixes: f044b318675f ("btrfs: handle the ro->rw transition for mounting different subvolumes")
Link: https://bugzilla.suse.com/show_bug.cgi?id=1231836
Signed-off-by: Qu Wenruo <wqu@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2024-10-30 11:25:48 +10:30
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2025-05-06 20:58:26 +01:00
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2023-11-22 12:17:48 -05:00
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2025-05-06 20:58:26 +01:00
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2023-11-22 12:17:48 -05:00
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2025-06-12 15:14:35 +09:30
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2023-11-22 12:17:48 -05:00
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2011-11-17 01:22:46 -05:00
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2011-11-17 15:40:49 -05:00
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2011-11-17 01:22:46 -05:00
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2020-01-24 09:32:53 -05:00
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2011-11-17 01:22:46 -05:00
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2023-11-22 12:17:46 -05:00
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2023-11-22 12:17:48 -05:00
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2023-11-22 12:17:46 -05:00
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2023-11-22 12:17:48 -05:00
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2023-11-22 12:17:46 -05:00
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2023-11-22 12:17:48 -05:00
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2023-11-22 12:17:46 -05:00
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2023-11-22 12:17:45 -05:00
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2023-11-22 12:17:47 -05:00
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2023-11-22 12:17:48 -05:00
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2023-11-22 12:17:46 -05:00
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2007-03-21 11:12:56 -04:00
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2017-12-14 17:24:30 +09:00
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2023-11-22 12:17:50 -05:00
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2023-11-22 12:17:46 -05:00
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2026-02-21 16:37:42 -08:00
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2023-11-22 12:17:46 -05:00
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2023-11-22 12:17:48 -05:00
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2023-11-22 12:17:46 -05:00
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2023-11-22 12:17:47 -05:00
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2023-11-22 12:17:50 -05:00
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2023-11-22 12:17:46 -05:00
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2007-03-21 11:12:56 -04:00
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2023-11-22 12:17:50 -05:00
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2024-10-02 17:27:26 -04:00
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2023-11-22 12:17:50 -05:00
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2017-12-14 17:24:30 +09:00
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2013-03-02 19:39:14 -08:00
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2008-03-24 15:02:04 -04:00
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2015-03-24 16:35:49 +01:00
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2008-09-29 15:18:18 -04:00
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2020-02-04 12:51:56 +08:00
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2008-09-29 15:18:18 -04:00
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2008-03-24 15:02:07 -04:00
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2018-07-12 14:23:16 +08:00
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2022-01-12 13:06:00 +08:00
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2009-01-16 11:59:08 -05:00
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2008-03-24 15:02:07 -04:00
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2009-01-05 16:57:23 -05:00
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2009-04-08 15:06:54 +08:00
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2024-02-14 10:32:47 +01:00
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2009-01-16 11:59:08 -05:00
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2008-03-24 15:02:07 -04:00
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2018-06-19 16:37:36 +02:00
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2023-09-09 00:31:55 +08:00
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2025-06-11 12:02:59 +02:00
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2018-07-12 14:23:16 +08:00
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2018-06-19 16:37:36 +02:00
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2008-03-24 15:02:07 -04:00
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2019-01-04 13:31:54 +08:00
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2022-01-12 13:06:00 +08:00
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2019-01-04 13:31:54 +08:00
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2012-06-21 16:03:58 -04:00
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2018-06-19 16:37:36 +02:00
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2023-09-09 00:31:55 +08:00
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2025-06-11 12:02:59 +02:00
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2023-09-09 00:31:55 +08:00
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2018-06-19 16:37:36 +02:00
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2025-08-12 16:25:54 +08:00
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2012-06-21 16:03:58 -04:00
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2018-06-19 16:37:36 +02:00
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2018-07-12 14:23:16 +08:00
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2018-06-19 16:37:36 +02:00
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2012-06-21 16:03:58 -04:00
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2016-02-17 15:24:14 +01:00
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2016-02-17 15:26:27 +01:00
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2016-02-17 15:24:14 +01:00
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2008-03-24 15:02:07 -04:00
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2009-04-08 15:06:54 +08:00
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2024-02-14 10:32:47 +01:00
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2008-03-24 15:02:07 -04:00
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2008-06-09 22:17:11 -04:00
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2008-03-24 15:02:07 -04:00
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2009-01-10 06:09:52 -08:00
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2008-01-22 12:46:56 -05:00
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2016-06-22 18:54:23 -04:00
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Btrfs: fix orphan transaction on the freezed filesystem
With the following debug patch:
static int btrfs_freeze(struct super_block *sb)
{
+ struct btrfs_fs_info *fs_info = btrfs_sb(sb);
+ struct btrfs_transaction *trans;
+
+ spin_lock(&fs_info->trans_lock);
+ trans = fs_info->running_transaction;
+ if (trans) {
+ printk("Transid %llu, use_count %d, num_writer %d\n",
+ trans->transid, atomic_read(&trans->use_count),
+ atomic_read(&trans->num_writers));
+ }
+ spin_unlock(&fs_info->trans_lock);
return 0;
}
I found there was a orphan transaction after the freeze operation was done.
It is because the transaction may not be committed when the transaction handle
end even though it is the last handle of the current transaction. This design
avoid committing the transaction frequently, but also introduce the above
problem.
So I add btrfs_attach_transaction() which can catch the current transaction
and commit it. If there is no transaction, it will return ENOENT, and do not
anything.
This function also can be used to instead of btrfs_join_transaction_freeze()
because it don't increase the writer counter and don't start a new transaction,
so it also can fix the deadlock between sync and freeze.
Besides that, it is used to instead of btrfs_join_transaction() in
transaction_kthread(), because if there is no transaction, the transaction
kthread needn't anything.
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
2012-09-20 01:54:00 -06:00
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2017-06-15 19:10:03 +02:00
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btrfs: fix fsfreeze hang caused by delayed iputs deal
When running fstests generic/068, sometimes we got below deadlock:
xfs_io D ffff8800331dbb20 0 6697 6693 0x00000080
ffff8800331dbb20 ffff88007acfc140 ffff880034d895c0 ffff8800331dc000
ffff880032d243e8 fffffffeffffffff ffff880032d24400 0000000000000001
ffff8800331dbb38 ffffffff816a9045 ffff880034d895c0 ffff8800331dbba8
Call Trace:
[<ffffffff816a9045>] schedule+0x35/0x80
[<ffffffff816abab2>] rwsem_down_read_failed+0xf2/0x140
[<ffffffff8118f5e1>] ? __filemap_fdatawrite_range+0xd1/0x100
[<ffffffff8134f978>] call_rwsem_down_read_failed+0x18/0x30
[<ffffffffa06631fc>] ? btrfs_alloc_block_rsv+0x2c/0xb0 [btrfs]
[<ffffffff810d32b5>] percpu_down_read+0x35/0x50
[<ffffffff81217dfc>] __sb_start_write+0x2c/0x40
[<ffffffffa067f5d5>] start_transaction+0x2a5/0x4d0 [btrfs]
[<ffffffffa067f857>] btrfs_join_transaction+0x17/0x20 [btrfs]
[<ffffffffa068ba34>] btrfs_evict_inode+0x3c4/0x5d0 [btrfs]
[<ffffffff81230a1a>] evict+0xba/0x1a0
[<ffffffff812316b6>] iput+0x196/0x200
[<ffffffffa06851d0>] btrfs_run_delayed_iputs+0x70/0xc0 [btrfs]
[<ffffffffa067f1d8>] btrfs_commit_transaction+0x928/0xa80 [btrfs]
[<ffffffffa0646df0>] btrfs_freeze+0x30/0x40 [btrfs]
[<ffffffff81218040>] freeze_super+0xf0/0x190
[<ffffffff81229275>] do_vfs_ioctl+0x4a5/0x5c0
[<ffffffff81003176>] ? do_audit_syscall_entry+0x66/0x70
[<ffffffff810038cf>] ? syscall_trace_enter_phase1+0x11f/0x140
[<ffffffff81229409>] SyS_ioctl+0x79/0x90
[<ffffffff81003c12>] do_syscall_64+0x62/0x110
[<ffffffff816acbe1>] entry_SYSCALL64_slow_path+0x25/0x25
>From this warning, freeze_super() already holds SB_FREEZE_FS, but
btrfs_freeze() will call btrfs_commit_transaction() again, if
btrfs_commit_transaction() finds that it has delayed iputs to handle,
it'll start_transaction(), which will try to get SB_FREEZE_FS lock
again, then deadlock occurs.
The root cause is that in btrfs, sync_filesystem(sb) does not make
sure all metadata is updated. There still maybe some codes adding
delayed iputs, see below sample race window:
CPU1 | CPU2
|-> freeze_super() |
|-> sync_filesystem(sb); |
| |-> cleaner_kthread()
| | |-> btrfs_delete_unused_bgs()
| | |-> btrfs_remove_chunk()
| | |-> btrfs_remove_block_group()
| | |-> btrfs_add_delayed_iput()
| |
|-> sb->s_writers.frozen = SB_FREEZE_FS; |
|-> sb_wait_write(sb, SB_FREEZE_FS); |
| acquire SB_FREEZE_FS lock. |
| |
|-> btrfs_freeze() |
|-> btrfs_commit_transaction() |
|-> btrfs_run_delayed_iputs() |
| will handle delayed iputs, |
| that means start_transaction() |
| will be called, which will try |
| to get SB_FREEZE_FS lock. |
To fix this issue, introduce a "int fs_frozen" to record internally whether
fs has been frozen. If fs has been frozen, we can not handle delayed iputs.
Signed-off-by: Wang Xiaoguang <wangxg.fnst@cn.fujitsu.com>
Reviewed-by: David Sterba <dsterba@suse.com>
[ add comment to btrfs_freeze ]
Signed-off-by: David Sterba <dsterba@suse.com>
Signed-off-by: Chris Mason <clm@fb.com>
2016-08-01 13:28:08 +08:00
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2024-05-22 09:26:44 +01:00
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2008-01-22 12:46:56 -05:00
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btrfs: check superblock to ensure the fs was not modified at thaw time
[BACKGROUND]
There is an incident report that, one user hibernated the system, with
one btrfs on removable device still mounted.
Then by some incident, the btrfs got mounted and modified by another
system/OS, then back to the hibernated system.
After resuming from the hibernation, new write happened into the victim btrfs.
Now the fs is completely broken, since the underlying btrfs is no longer
the same one before the hibernation, and the user lost their data due to
various transid mismatch.
[REPRODUCER]
We can emulate the situation using the following small script:
truncate -s 1G $dev
mkfs.btrfs -f $dev
mount $dev $mnt
fsstress -w -d $mnt -n 500
sync
xfs_freeze -f $mnt
cp $dev $dev.backup
# There is no way to mount the same cloned fs on the same system,
# as the conflicting fsid will be rejected by btrfs.
# Thus here we have to wipe the fs using a different btrfs.
mkfs.btrfs -f $dev.backup
dd if=$dev.backup of=$dev bs=1M
xfs_freeze -u $mnt
fsstress -w -d $mnt -n 20
umount $mnt
btrfs check $dev
The final fsck will fail due to some tree blocks has incorrect fsid.
This is enough to emulate the problem hit by the unfortunate user.
[ENHANCEMENT]
Although such case should not be that common, it can still happen from
time to time.
From the view of btrfs, we can detect any unexpected super block change,
and if there is any unexpected change, we just mark the fs read-only,
and thaw the fs.
By this we can limit the damage to minimal, and I hope no one would lose
their data by this anymore.
Suggested-by: Goffredo Baroncelli <kreijack@libero.it>
Link: https://lore.kernel.org/linux-btrfs/83bf3b4b-7f4c-387a-b286-9251e3991e34@bluemole.com/
Reviewed-by: Anand Jain <anand.jain@oracle.com>
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2022-08-24 20:16:22 +08:00
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2023-10-04 11:38:51 +01:00
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2022-10-18 09:56:38 +08:00
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|
btrfs: check superblock to ensure the fs was not modified at thaw time
[BACKGROUND]
There is an incident report that, one user hibernated the system, with
one btrfs on removable device still mounted.
Then by some incident, the btrfs got mounted and modified by another
system/OS, then back to the hibernated system.
After resuming from the hibernation, new write happened into the victim btrfs.
Now the fs is completely broken, since the underlying btrfs is no longer
the same one before the hibernation, and the user lost their data due to
various transid mismatch.
[REPRODUCER]
We can emulate the situation using the following small script:
truncate -s 1G $dev
mkfs.btrfs -f $dev
mount $dev $mnt
fsstress -w -d $mnt -n 500
sync
xfs_freeze -f $mnt
cp $dev $dev.backup
# There is no way to mount the same cloned fs on the same system,
# as the conflicting fsid will be rejected by btrfs.
# Thus here we have to wipe the fs using a different btrfs.
mkfs.btrfs -f $dev.backup
dd if=$dev.backup of=$dev bs=1M
xfs_freeze -u $mnt
fsstress -w -d $mnt -n 20
umount $mnt
btrfs check $dev
The final fsck will fail due to some tree blocks has incorrect fsid.
This is enough to emulate the problem hit by the unfortunate user.
[ENHANCEMENT]
Although such case should not be that common, it can still happen from
time to time.
From the view of btrfs, we can detect any unexpected super block change,
and if there is any unexpected change, we just mark the fs read-only,
and thaw the fs.
By this we can limit the damage to minimal, and I hope no one would lose
their data by this anymore.
Suggested-by: Goffredo Baroncelli <kreijack@libero.it>
Link: https://lore.kernel.org/linux-btrfs/83bf3b4b-7f4c-387a-b286-9251e3991e34@bluemole.com/
Reviewed-by: Anand Jain <anand.jain@oracle.com>
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2022-08-24 20:16:22 +08:00
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2025-04-28 10:06:50 +09:30
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|
btrfs: check superblock to ensure the fs was not modified at thaw time
[BACKGROUND]
There is an incident report that, one user hibernated the system, with
one btrfs on removable device still mounted.
Then by some incident, the btrfs got mounted and modified by another
system/OS, then back to the hibernated system.
After resuming from the hibernation, new write happened into the victim btrfs.
Now the fs is completely broken, since the underlying btrfs is no longer
the same one before the hibernation, and the user lost their data due to
various transid mismatch.
[REPRODUCER]
We can emulate the situation using the following small script:
truncate -s 1G $dev
mkfs.btrfs -f $dev
mount $dev $mnt
fsstress -w -d $mnt -n 500
sync
xfs_freeze -f $mnt
cp $dev $dev.backup
# There is no way to mount the same cloned fs on the same system,
# as the conflicting fsid will be rejected by btrfs.
# Thus here we have to wipe the fs using a different btrfs.
mkfs.btrfs -f $dev.backup
dd if=$dev.backup of=$dev bs=1M
xfs_freeze -u $mnt
fsstress -w -d $mnt -n 20
umount $mnt
btrfs check $dev
The final fsck will fail due to some tree blocks has incorrect fsid.
This is enough to emulate the problem hit by the unfortunate user.
[ENHANCEMENT]
Although such case should not be that common, it can still happen from
time to time.
From the view of btrfs, we can detect any unexpected super block change,
and if there is any unexpected change, we just mark the fs read-only,
and thaw the fs.
By this we can limit the damage to minimal, and I hope no one would lose
their data by this anymore.
Suggested-by: Goffredo Baroncelli <kreijack@libero.it>
Link: https://lore.kernel.org/linux-btrfs/83bf3b4b-7f4c-387a-b286-9251e3991e34@bluemole.com/
Reviewed-by: Anand Jain <anand.jain@oracle.com>
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2022-08-24 20:16:22 +08:00
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2022-10-18 09:56:38 +08:00
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2025-09-17 19:53:54 +02:00
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2022-10-18 09:56:38 +08:00
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2025-09-17 19:53:54 +02:00
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2022-10-18 09:56:38 +08:00
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btrfs: check superblock to ensure the fs was not modified at thaw time
[BACKGROUND]
There is an incident report that, one user hibernated the system, with
one btrfs on removable device still mounted.
Then by some incident, the btrfs got mounted and modified by another
system/OS, then back to the hibernated system.
After resuming from the hibernation, new write happened into the victim btrfs.
Now the fs is completely broken, since the underlying btrfs is no longer
the same one before the hibernation, and the user lost their data due to
various transid mismatch.
[REPRODUCER]
We can emulate the situation using the following small script:
truncate -s 1G $dev
mkfs.btrfs -f $dev
mount $dev $mnt
fsstress -w -d $mnt -n 500
sync
xfs_freeze -f $mnt
cp $dev $dev.backup
# There is no way to mount the same cloned fs on the same system,
# as the conflicting fsid will be rejected by btrfs.
# Thus here we have to wipe the fs using a different btrfs.
mkfs.btrfs -f $dev.backup
dd if=$dev.backup of=$dev bs=1M
xfs_freeze -u $mnt
fsstress -w -d $mnt -n 20
umount $mnt
btrfs check $dev
The final fsck will fail due to some tree blocks has incorrect fsid.
This is enough to emulate the problem hit by the unfortunate user.
[ENHANCEMENT]
Although such case should not be that common, it can still happen from
time to time.
From the view of btrfs, we can detect any unexpected super block change,
and if there is any unexpected change, we just mark the fs read-only,
and thaw the fs.
By this we can limit the damage to minimal, and I hope no one would lose
their data by this anymore.
Suggested-by: Goffredo Baroncelli <kreijack@libero.it>
Link: https://lore.kernel.org/linux-btrfs/83bf3b4b-7f4c-387a-b286-9251e3991e34@bluemole.com/
Reviewed-by: Anand Jain <anand.jain@oracle.com>
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2022-08-24 20:16:22 +08:00
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2023-10-04 11:38:51 +01:00
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2025-09-17 19:53:54 +02:00
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btrfs: check superblock to ensure the fs was not modified at thaw time
[BACKGROUND]
There is an incident report that, one user hibernated the system, with
one btrfs on removable device still mounted.
Then by some incident, the btrfs got mounted and modified by another
system/OS, then back to the hibernated system.
After resuming from the hibernation, new write happened into the victim btrfs.
Now the fs is completely broken, since the underlying btrfs is no longer
the same one before the hibernation, and the user lost their data due to
various transid mismatch.
[REPRODUCER]
We can emulate the situation using the following small script:
truncate -s 1G $dev
mkfs.btrfs -f $dev
mount $dev $mnt
fsstress -w -d $mnt -n 500
sync
xfs_freeze -f $mnt
cp $dev $dev.backup
# There is no way to mount the same cloned fs on the same system,
# as the conflicting fsid will be rejected by btrfs.
# Thus here we have to wipe the fs using a different btrfs.
mkfs.btrfs -f $dev.backup
dd if=$dev.backup of=$dev bs=1M
xfs_freeze -u $mnt
fsstress -w -d $mnt -n 20
umount $mnt
btrfs check $dev
The final fsck will fail due to some tree blocks has incorrect fsid.
This is enough to emulate the problem hit by the unfortunate user.
[ENHANCEMENT]
Although such case should not be that common, it can still happen from
time to time.
From the view of btrfs, we can detect any unexpected super block change,
and if there is any unexpected change, we just mark the fs read-only,
and thaw the fs.
By this we can limit the damage to minimal, and I hope no one would lose
their data by this anymore.
Suggested-by: Goffredo Baroncelli <kreijack@libero.it>
Link: https://lore.kernel.org/linux-btrfs/83bf3b4b-7f4c-387a-b286-9251e3991e34@bluemole.com/
Reviewed-by: Anand Jain <anand.jain@oracle.com>
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2022-08-24 20:16:22 +08:00
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2023-10-04 11:38:51 +01:00
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btrfs: check superblock to ensure the fs was not modified at thaw time
[BACKGROUND]
There is an incident report that, one user hibernated the system, with
one btrfs on removable device still mounted.
Then by some incident, the btrfs got mounted and modified by another
system/OS, then back to the hibernated system.
After resuming from the hibernation, new write happened into the victim btrfs.
Now the fs is completely broken, since the underlying btrfs is no longer
the same one before the hibernation, and the user lost their data due to
various transid mismatch.
[REPRODUCER]
We can emulate the situation using the following small script:
truncate -s 1G $dev
mkfs.btrfs -f $dev
mount $dev $mnt
fsstress -w -d $mnt -n 500
sync
xfs_freeze -f $mnt
cp $dev $dev.backup
# There is no way to mount the same cloned fs on the same system,
# as the conflicting fsid will be rejected by btrfs.
# Thus here we have to wipe the fs using a different btrfs.
mkfs.btrfs -f $dev.backup
dd if=$dev.backup of=$dev bs=1M
xfs_freeze -u $mnt
fsstress -w -d $mnt -n 20
umount $mnt
btrfs check $dev
The final fsck will fail due to some tree blocks has incorrect fsid.
This is enough to emulate the problem hit by the unfortunate user.
[ENHANCEMENT]
Although such case should not be that common, it can still happen from
time to time.
From the view of btrfs, we can detect any unexpected super block change,
and if there is any unexpected change, we just mark the fs read-only,
and thaw the fs.
By this we can limit the damage to minimal, and I hope no one would lose
their data by this anymore.
Suggested-by: Goffredo Baroncelli <kreijack@libero.it>
Link: https://lore.kernel.org/linux-btrfs/83bf3b4b-7f4c-387a-b286-9251e3991e34@bluemole.com/
Reviewed-by: Anand Jain <anand.jain@oracle.com>
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2022-08-24 20:16:22 +08:00
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btrfs: fix fsfreeze hang caused by delayed iputs deal
When running fstests generic/068, sometimes we got below deadlock:
xfs_io D ffff8800331dbb20 0 6697 6693 0x00000080
ffff8800331dbb20 ffff88007acfc140 ffff880034d895c0 ffff8800331dc000
ffff880032d243e8 fffffffeffffffff ffff880032d24400 0000000000000001
ffff8800331dbb38 ffffffff816a9045 ffff880034d895c0 ffff8800331dbba8
Call Trace:
[<ffffffff816a9045>] schedule+0x35/0x80
[<ffffffff816abab2>] rwsem_down_read_failed+0xf2/0x140
[<ffffffff8118f5e1>] ? __filemap_fdatawrite_range+0xd1/0x100
[<ffffffff8134f978>] call_rwsem_down_read_failed+0x18/0x30
[<ffffffffa06631fc>] ? btrfs_alloc_block_rsv+0x2c/0xb0 [btrfs]
[<ffffffff810d32b5>] percpu_down_read+0x35/0x50
[<ffffffff81217dfc>] __sb_start_write+0x2c/0x40
[<ffffffffa067f5d5>] start_transaction+0x2a5/0x4d0 [btrfs]
[<ffffffffa067f857>] btrfs_join_transaction+0x17/0x20 [btrfs]
[<ffffffffa068ba34>] btrfs_evict_inode+0x3c4/0x5d0 [btrfs]
[<ffffffff81230a1a>] evict+0xba/0x1a0
[<ffffffff812316b6>] iput+0x196/0x200
[<ffffffffa06851d0>] btrfs_run_delayed_iputs+0x70/0xc0 [btrfs]
[<ffffffffa067f1d8>] btrfs_commit_transaction+0x928/0xa80 [btrfs]
[<ffffffffa0646df0>] btrfs_freeze+0x30/0x40 [btrfs]
[<ffffffff81218040>] freeze_super+0xf0/0x190
[<ffffffff81229275>] do_vfs_ioctl+0x4a5/0x5c0
[<ffffffff81003176>] ? do_audit_syscall_entry+0x66/0x70
[<ffffffff810038cf>] ? syscall_trace_enter_phase1+0x11f/0x140
[<ffffffff81229409>] SyS_ioctl+0x79/0x90
[<ffffffff81003c12>] do_syscall_64+0x62/0x110
[<ffffffff816acbe1>] entry_SYSCALL64_slow_path+0x25/0x25
>From this warning, freeze_super() already holds SB_FREEZE_FS, but
btrfs_freeze() will call btrfs_commit_transaction() again, if
btrfs_commit_transaction() finds that it has delayed iputs to handle,
it'll start_transaction(), which will try to get SB_FREEZE_FS lock
again, then deadlock occurs.
The root cause is that in btrfs, sync_filesystem(sb) does not make
sure all metadata is updated. There still maybe some codes adding
delayed iputs, see below sample race window:
CPU1 | CPU2
|-> freeze_super() |
|-> sync_filesystem(sb); |
| |-> cleaner_kthread()
| | |-> btrfs_delete_unused_bgs()
| | |-> btrfs_remove_chunk()
| | |-> btrfs_remove_block_group()
| | |-> btrfs_add_delayed_iput()
| |
|-> sb->s_writers.frozen = SB_FREEZE_FS; |
|-> sb_wait_write(sb, SB_FREEZE_FS); |
| acquire SB_FREEZE_FS lock. |
| |
|-> btrfs_freeze() |
|-> btrfs_commit_transaction() |
|-> btrfs_run_delayed_iputs() |
| will handle delayed iputs, |
| that means start_transaction() |
| will be called, which will try |
| to get SB_FREEZE_FS lock. |
To fix this issue, introduce a "int fs_frozen" to record internally whether
fs has been frozen. If fs has been frozen, we can not handle delayed iputs.
Signed-off-by: Wang Xiaoguang <wangxg.fnst@cn.fujitsu.com>
Reviewed-by: David Sterba <dsterba@suse.com>
[ add comment to btrfs_freeze ]
Signed-off-by: David Sterba <dsterba@suse.com>
Signed-off-by: Chris Mason <clm@fb.com>
2016-08-01 13:28:08 +08:00
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2017-06-15 19:10:03 +02:00
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btrfs: check superblock to ensure the fs was not modified at thaw time
[BACKGROUND]
There is an incident report that, one user hibernated the system, with
one btrfs on removable device still mounted.
Then by some incident, the btrfs got mounted and modified by another
system/OS, then back to the hibernated system.
After resuming from the hibernation, new write happened into the victim btrfs.
Now the fs is completely broken, since the underlying btrfs is no longer
the same one before the hibernation, and the user lost their data due to
various transid mismatch.
[REPRODUCER]
We can emulate the situation using the following small script:
truncate -s 1G $dev
mkfs.btrfs -f $dev
mount $dev $mnt
fsstress -w -d $mnt -n 500
sync
xfs_freeze -f $mnt
cp $dev $dev.backup
# There is no way to mount the same cloned fs on the same system,
# as the conflicting fsid will be rejected by btrfs.
# Thus here we have to wipe the fs using a different btrfs.
mkfs.btrfs -f $dev.backup
dd if=$dev.backup of=$dev bs=1M
xfs_freeze -u $mnt
fsstress -w -d $mnt -n 20
umount $mnt
btrfs check $dev
The final fsck will fail due to some tree blocks has incorrect fsid.
This is enough to emulate the problem hit by the unfortunate user.
[ENHANCEMENT]
Although such case should not be that common, it can still happen from
time to time.
From the view of btrfs, we can detect any unexpected super block change,
and if there is any unexpected change, we just mark the fs read-only,
and thaw the fs.
By this we can limit the damage to minimal, and I hope no one would lose
their data by this anymore.
Suggested-by: Goffredo Baroncelli <kreijack@libero.it>
Link: https://lore.kernel.org/linux-btrfs/83bf3b4b-7f4c-387a-b286-9251e3991e34@bluemole.com/
Reviewed-by: Anand Jain <anand.jain@oracle.com>
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2022-08-24 20:16:22 +08:00
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2017-06-15 19:10:03 +02:00
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|
btrfs: check superblock to ensure the fs was not modified at thaw time
[BACKGROUND]
There is an incident report that, one user hibernated the system, with
one btrfs on removable device still mounted.
Then by some incident, the btrfs got mounted and modified by another
system/OS, then back to the hibernated system.
After resuming from the hibernation, new write happened into the victim btrfs.
Now the fs is completely broken, since the underlying btrfs is no longer
the same one before the hibernation, and the user lost their data due to
various transid mismatch.
[REPRODUCER]
We can emulate the situation using the following small script:
truncate -s 1G $dev
mkfs.btrfs -f $dev
mount $dev $mnt
fsstress -w -d $mnt -n 500
sync
xfs_freeze -f $mnt
cp $dev $dev.backup
# There is no way to mount the same cloned fs on the same system,
# as the conflicting fsid will be rejected by btrfs.
# Thus here we have to wipe the fs using a different btrfs.
mkfs.btrfs -f $dev.backup
dd if=$dev.backup of=$dev bs=1M
xfs_freeze -u $mnt
fsstress -w -d $mnt -n 20
umount $mnt
btrfs check $dev
The final fsck will fail due to some tree blocks has incorrect fsid.
This is enough to emulate the problem hit by the unfortunate user.
[ENHANCEMENT]
Although such case should not be that common, it can still happen from
time to time.
From the view of btrfs, we can detect any unexpected super block change,
and if there is any unexpected change, we just mark the fs read-only,
and thaw the fs.
By this we can limit the damage to minimal, and I hope no one would lose
their data by this anymore.
Suggested-by: Goffredo Baroncelli <kreijack@libero.it>
Link: https://lore.kernel.org/linux-btrfs/83bf3b4b-7f4c-387a-b286-9251e3991e34@bluemole.com/
Reviewed-by: Anand Jain <anand.jain@oracle.com>
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2022-08-24 20:16:22 +08:00
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2017-06-15 19:10:03 +02:00
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btrfs: check superblock to ensure the fs was not modified at thaw time
[BACKGROUND]
There is an incident report that, one user hibernated the system, with
one btrfs on removable device still mounted.
Then by some incident, the btrfs got mounted and modified by another
system/OS, then back to the hibernated system.
After resuming from the hibernation, new write happened into the victim btrfs.
Now the fs is completely broken, since the underlying btrfs is no longer
the same one before the hibernation, and the user lost their data due to
various transid mismatch.
[REPRODUCER]
We can emulate the situation using the following small script:
truncate -s 1G $dev
mkfs.btrfs -f $dev
mount $dev $mnt
fsstress -w -d $mnt -n 500
sync
xfs_freeze -f $mnt
cp $dev $dev.backup
# There is no way to mount the same cloned fs on the same system,
# as the conflicting fsid will be rejected by btrfs.
# Thus here we have to wipe the fs using a different btrfs.
mkfs.btrfs -f $dev.backup
dd if=$dev.backup of=$dev bs=1M
xfs_freeze -u $mnt
fsstress -w -d $mnt -n 20
umount $mnt
btrfs check $dev
The final fsck will fail due to some tree blocks has incorrect fsid.
This is enough to emulate the problem hit by the unfortunate user.
[ENHANCEMENT]
Although such case should not be that common, it can still happen from
time to time.
From the view of btrfs, we can detect any unexpected super block change,
and if there is any unexpected change, we just mark the fs read-only,
and thaw the fs.
By this we can limit the damage to minimal, and I hope no one would lose
their data by this anymore.
Suggested-by: Goffredo Baroncelli <kreijack@libero.it>
Link: https://lore.kernel.org/linux-btrfs/83bf3b4b-7f4c-387a-b286-9251e3991e34@bluemole.com/
Reviewed-by: Anand Jain <anand.jain@oracle.com>
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2022-08-24 20:16:22 +08:00
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btrfs: fix fsfreeze hang caused by delayed iputs deal
When running fstests generic/068, sometimes we got below deadlock:
xfs_io D ffff8800331dbb20 0 6697 6693 0x00000080
ffff8800331dbb20 ffff88007acfc140 ffff880034d895c0 ffff8800331dc000
ffff880032d243e8 fffffffeffffffff ffff880032d24400 0000000000000001
ffff8800331dbb38 ffffffff816a9045 ffff880034d895c0 ffff8800331dbba8
Call Trace:
[<ffffffff816a9045>] schedule+0x35/0x80
[<ffffffff816abab2>] rwsem_down_read_failed+0xf2/0x140
[<ffffffff8118f5e1>] ? __filemap_fdatawrite_range+0xd1/0x100
[<ffffffff8134f978>] call_rwsem_down_read_failed+0x18/0x30
[<ffffffffa06631fc>] ? btrfs_alloc_block_rsv+0x2c/0xb0 [btrfs]
[<ffffffff810d32b5>] percpu_down_read+0x35/0x50
[<ffffffff81217dfc>] __sb_start_write+0x2c/0x40
[<ffffffffa067f5d5>] start_transaction+0x2a5/0x4d0 [btrfs]
[<ffffffffa067f857>] btrfs_join_transaction+0x17/0x20 [btrfs]
[<ffffffffa068ba34>] btrfs_evict_inode+0x3c4/0x5d0 [btrfs]
[<ffffffff81230a1a>] evict+0xba/0x1a0
[<ffffffff812316b6>] iput+0x196/0x200
[<ffffffffa06851d0>] btrfs_run_delayed_iputs+0x70/0xc0 [btrfs]
[<ffffffffa067f1d8>] btrfs_commit_transaction+0x928/0xa80 [btrfs]
[<ffffffffa0646df0>] btrfs_freeze+0x30/0x40 [btrfs]
[<ffffffff81218040>] freeze_super+0xf0/0x190
[<ffffffff81229275>] do_vfs_ioctl+0x4a5/0x5c0
[<ffffffff81003176>] ? do_audit_syscall_entry+0x66/0x70
[<ffffffff810038cf>] ? syscall_trace_enter_phase1+0x11f/0x140
[<ffffffff81229409>] SyS_ioctl+0x79/0x90
[<ffffffff81003c12>] do_syscall_64+0x62/0x110
[<ffffffff816acbe1>] entry_SYSCALL64_slow_path+0x25/0x25
>From this warning, freeze_super() already holds SB_FREEZE_FS, but
btrfs_freeze() will call btrfs_commit_transaction() again, if
btrfs_commit_transaction() finds that it has delayed iputs to handle,
it'll start_transaction(), which will try to get SB_FREEZE_FS lock
again, then deadlock occurs.
The root cause is that in btrfs, sync_filesystem(sb) does not make
sure all metadata is updated. There still maybe some codes adding
delayed iputs, see below sample race window:
CPU1 | CPU2
|-> freeze_super() |
|-> sync_filesystem(sb); |
| |-> cleaner_kthread()
| | |-> btrfs_delete_unused_bgs()
| | |-> btrfs_remove_chunk()
| | |-> btrfs_remove_block_group()
| | |-> btrfs_add_delayed_iput()
| |
|-> sb->s_writers.frozen = SB_FREEZE_FS; |
|-> sb_wait_write(sb, SB_FREEZE_FS); |
| acquire SB_FREEZE_FS lock. |
| |
|-> btrfs_freeze() |
|-> btrfs_commit_transaction() |
|-> btrfs_run_delayed_iputs() |
| will handle delayed iputs, |
| that means start_transaction() |
| will be called, which will try |
| to get SB_FREEZE_FS lock. |
To fix this issue, introduce a "int fs_frozen" to record internally whether
fs has been frozen. If fs has been frozen, we can not handle delayed iputs.
Signed-off-by: Wang Xiaoguang <wangxg.fnst@cn.fujitsu.com>
Reviewed-by: David Sterba <dsterba@suse.com>
[ add comment to btrfs_freeze ]
Signed-off-by: David Sterba <dsterba@suse.com>
Signed-off-by: Chris Mason <clm@fb.com>
2016-08-01 13:28:08 +08:00
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2012-06-05 14:13:12 -04:00
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2018-03-16 03:27:02 +01:00
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2021-08-24 13:05:20 +08:00
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2018-03-16 03:27:02 +01:00
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2022-11-13 09:32:07 +08:00
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2018-03-16 03:27:02 +01:00
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2021-08-24 13:05:20 +08:00
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2012-06-05 14:13:12 -04:00
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btrfs: add a shrinker for extent maps
Extent maps are used either to represent existing file extent items, or to
represent new extents that are going to be written and the respective file
extent items are created when the ordered extent completes.
We currently don't have any limit for how many extent maps we can have,
neither per inode nor globally. Most of the time this not too noticeable
because extent maps are removed in the following situations:
1) When evicting an inode;
2) When releasing folios (pages) through the btrfs_release_folio() address
space operation callback.
However we won't release extent maps in the folio range if the folio is
either dirty or under writeback or if the inode's i_size is less than
or equals to 16M (see try_release_extent_mapping(). This 16M i_size
constraint was added back in 2008 with commit 70dec8079d78 ("Btrfs:
extent_io and extent_state optimizations"), but there's no explanation
about why we have it or why the 16M value.
This means that for buffered IO we can reach an OOM situation due to too
many extent maps if either of the following happens:
1) There's a set of tasks constantly doing IO on many files with a size
not larger than 16M, specially if they keep the files open for very
long periods, therefore preventing inode eviction.
This requires a really high number of such files, and having many non
mergeable extent maps (due to random 4K writes for example) and a
machine with very little memory;
2) There's a set tasks constantly doing random write IO (therefore
creating many non mergeable extent maps) on files and keeping them
open for long periods of time, so inode eviction doesn't happen and
there's always a lot of dirty pages or pages under writeback,
preventing btrfs_release_folio() from releasing the respective extent
maps.
This second case was actually reported in the thread pointed by the Link
tag below, and it requires a very large file under heavy IO and a machine
with very little amount of RAM, which is probably hard to happen in
practice in a real world use case.
However when using direct IO this is not so hard to happen, because the
page cache is not used, and therefore btrfs_release_folio() is never
called. Which means extent maps are dropped only when evicting the inode,
and that means that if we have tasks that keep a file descriptor open and
keep doing IO on a very large file (or files), we can exhaust memory due
to an unbounded amount of extent maps. This is especially easy to happen
if we have a huge file with millions of small extents and their extent
maps are not mergeable (non contiguous offsets and disk locations).
This was reported in that thread with the following fio test:
$ cat test.sh
#!/bin/bash
DEV=/dev/sdj
MNT=/mnt/sdj
MOUNT_OPTIONS="-o ssd"
MKFS_OPTIONS=""
cat <<EOF > /tmp/fio-job.ini
[global]
name=fio-rand-write
filename=$MNT/fio-rand-write
rw=randwrite
bs=4K
direct=1
numjobs=16
fallocate=none
time_based
runtime=90000
[file1]
size=300G
ioengine=libaio
iodepth=16
EOF
umount $MNT &> /dev/null
mkfs.btrfs -f $MKFS_OPTIONS $DEV
mount $MOUNT_OPTIONS $DEV $MNT
fio /tmp/fio-job.ini
umount $MNT
Monitoring the btrfs_extent_map slab while running the test with:
$ watch -d -n 1 'cat /sys/kernel/slab/btrfs_extent_map/objects \
/sys/kernel/slab/btrfs_extent_map/total_objects'
Shows the number of active and total extent maps skyrocketing to tens of
millions, and on systems with a short amount of memory it's easy and quick
to get into an OOM situation, as reported in that thread.
So to avoid this issue add a shrinker that will remove extents maps, as
long as they are not pinned, and takes proper care with any concurrent
fsync to avoid missing extents (setting the full sync flag while in the
middle of a fast fsync). This shrinker is triggered through the callbacks
nr_cached_objects and free_cached_objects of struct super_operations.
The shrinker will iterate over all roots and over all inodes of each
root, and keeps track of the last scanned root and inode, so that the
next time it runs, it starts from that root and from the next inode.
This is similar to what xfs does for its inode reclaim (implements those
callbacks, and cycles through inodes by starting from where it ended
last time).
Reviewed-by: Josef Bacik <josef@toxicpanda.com>
Signed-off-by: Filipe Manana <fdmanana@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2024-04-15 17:09:26 +01:00
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2024-04-09 17:07:32 +01:00
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btrfs: add a shrinker for extent maps
Extent maps are used either to represent existing file extent items, or to
represent new extents that are going to be written and the respective file
extent items are created when the ordered extent completes.
We currently don't have any limit for how many extent maps we can have,
neither per inode nor globally. Most of the time this not too noticeable
because extent maps are removed in the following situations:
1) When evicting an inode;
2) When releasing folios (pages) through the btrfs_release_folio() address
space operation callback.
However we won't release extent maps in the folio range if the folio is
either dirty or under writeback or if the inode's i_size is less than
or equals to 16M (see try_release_extent_mapping(). This 16M i_size
constraint was added back in 2008 with commit 70dec8079d78 ("Btrfs:
extent_io and extent_state optimizations"), but there's no explanation
about why we have it or why the 16M value.
This means that for buffered IO we can reach an OOM situation due to too
many extent maps if either of the following happens:
1) There's a set of tasks constantly doing IO on many files with a size
not larger than 16M, specially if they keep the files open for very
long periods, therefore preventing inode eviction.
This requires a really high number of such files, and having many non
mergeable extent maps (due to random 4K writes for example) and a
machine with very little memory;
2) There's a set tasks constantly doing random write IO (therefore
creating many non mergeable extent maps) on files and keeping them
open for long periods of time, so inode eviction doesn't happen and
there's always a lot of dirty pages or pages under writeback,
preventing btrfs_release_folio() from releasing the respective extent
maps.
This second case was actually reported in the thread pointed by the Link
tag below, and it requires a very large file under heavy IO and a machine
with very little amount of RAM, which is probably hard to happen in
practice in a real world use case.
However when using direct IO this is not so hard to happen, because the
page cache is not used, and therefore btrfs_release_folio() is never
called. Which means extent maps are dropped only when evicting the inode,
and that means that if we have tasks that keep a file descriptor open and
keep doing IO on a very large file (or files), we can exhaust memory due
to an unbounded amount of extent maps. This is especially easy to happen
if we have a huge file with millions of small extents and their extent
maps are not mergeable (non contiguous offsets and disk locations).
This was reported in that thread with the following fio test:
$ cat test.sh
#!/bin/bash
DEV=/dev/sdj
MNT=/mnt/sdj
MOUNT_OPTIONS="-o ssd"
MKFS_OPTIONS=""
cat <<EOF > /tmp/fio-job.ini
[global]
name=fio-rand-write
filename=$MNT/fio-rand-write
rw=randwrite
bs=4K
direct=1
numjobs=16
fallocate=none
time_based
runtime=90000
[file1]
size=300G
ioengine=libaio
iodepth=16
EOF
umount $MNT &> /dev/null
mkfs.btrfs -f $MKFS_OPTIONS $DEV
mount $MOUNT_OPTIONS $DEV $MNT
fio /tmp/fio-job.ini
umount $MNT
Monitoring the btrfs_extent_map slab while running the test with:
$ watch -d -n 1 'cat /sys/kernel/slab/btrfs_extent_map/objects \
/sys/kernel/slab/btrfs_extent_map/total_objects'
Shows the number of active and total extent maps skyrocketing to tens of
millions, and on systems with a short amount of memory it's easy and quick
to get into an OOM situation, as reported in that thread.
So to avoid this issue add a shrinker that will remove extents maps, as
long as they are not pinned, and takes proper care with any concurrent
fsync to avoid missing extents (setting the full sync flag while in the
middle of a fast fsync). This shrinker is triggered through the callbacks
nr_cached_objects and free_cached_objects of struct super_operations.
The shrinker will iterate over all roots and over all inodes of each
root, and keeps track of the last scanned root and inode, so that the
next time it runs, it starts from that root and from the next inode.
This is similar to what xfs does for its inode reclaim (implements those
callbacks, and cycles through inodes by starting from where it ended
last time).
Reviewed-by: Josef Bacik <josef@toxicpanda.com>
Signed-off-by: Filipe Manana <fdmanana@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2024-04-15 17:09:26 +01:00
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2024-04-09 17:07:32 +01:00
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2024-09-17 12:03:11 +01:00
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btrfs: add a shrinker for extent maps
Extent maps are used either to represent existing file extent items, or to
represent new extents that are going to be written and the respective file
extent items are created when the ordered extent completes.
We currently don't have any limit for how many extent maps we can have,
neither per inode nor globally. Most of the time this not too noticeable
because extent maps are removed in the following situations:
1) When evicting an inode;
2) When releasing folios (pages) through the btrfs_release_folio() address
space operation callback.
However we won't release extent maps in the folio range if the folio is
either dirty or under writeback or if the inode's i_size is less than
or equals to 16M (see try_release_extent_mapping(). This 16M i_size
constraint was added back in 2008 with commit 70dec8079d78 ("Btrfs:
extent_io and extent_state optimizations"), but there's no explanation
about why we have it or why the 16M value.
This means that for buffered IO we can reach an OOM situation due to too
many extent maps if either of the following happens:
1) There's a set of tasks constantly doing IO on many files with a size
not larger than 16M, specially if they keep the files open for very
long periods, therefore preventing inode eviction.
This requires a really high number of such files, and having many non
mergeable extent maps (due to random 4K writes for example) and a
machine with very little memory;
2) There's a set tasks constantly doing random write IO (therefore
creating many non mergeable extent maps) on files and keeping them
open for long periods of time, so inode eviction doesn't happen and
there's always a lot of dirty pages or pages under writeback,
preventing btrfs_release_folio() from releasing the respective extent
maps.
This second case was actually reported in the thread pointed by the Link
tag below, and it requires a very large file under heavy IO and a machine
with very little amount of RAM, which is probably hard to happen in
practice in a real world use case.
However when using direct IO this is not so hard to happen, because the
page cache is not used, and therefore btrfs_release_folio() is never
called. Which means extent maps are dropped only when evicting the inode,
and that means that if we have tasks that keep a file descriptor open and
keep doing IO on a very large file (or files), we can exhaust memory due
to an unbounded amount of extent maps. This is especially easy to happen
if we have a huge file with millions of small extents and their extent
maps are not mergeable (non contiguous offsets and disk locations).
This was reported in that thread with the following fio test:
$ cat test.sh
#!/bin/bash
DEV=/dev/sdj
MNT=/mnt/sdj
MOUNT_OPTIONS="-o ssd"
MKFS_OPTIONS=""
cat <<EOF > /tmp/fio-job.ini
[global]
name=fio-rand-write
filename=$MNT/fio-rand-write
rw=randwrite
bs=4K
direct=1
numjobs=16
fallocate=none
time_based
runtime=90000
[file1]
size=300G
ioengine=libaio
iodepth=16
EOF
umount $MNT &> /dev/null
mkfs.btrfs -f $MKFS_OPTIONS $DEV
mount $MOUNT_OPTIONS $DEV $MNT
fio /tmp/fio-job.ini
umount $MNT
Monitoring the btrfs_extent_map slab while running the test with:
$ watch -d -n 1 'cat /sys/kernel/slab/btrfs_extent_map/objects \
/sys/kernel/slab/btrfs_extent_map/total_objects'
Shows the number of active and total extent maps skyrocketing to tens of
millions, and on systems with a short amount of memory it's easy and quick
to get into an OOM situation, as reported in that thread.
So to avoid this issue add a shrinker that will remove extents maps, as
long as they are not pinned, and takes proper care with any concurrent
fsync to avoid missing extents (setting the full sync flag while in the
middle of a fast fsync). This shrinker is triggered through the callbacks
nr_cached_objects and free_cached_objects of struct super_operations.
The shrinker will iterate over all roots and over all inodes of each
root, and keeps track of the last scanned root and inode, so that the
next time it runs, it starts from that root and from the next inode.
This is similar to what xfs does for its inode reclaim (implements those
callbacks, and cycles through inodes by starting from where it ended
last time).
Reviewed-by: Josef Bacik <josef@toxicpanda.com>
Signed-off-by: Filipe Manana <fdmanana@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2024-04-15 17:09:26 +01:00
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btrfs: make the extent map shrinker run asynchronously as a work queue job
Currently the extent map shrinker is run synchronously for kswapd tasks
that end up calling the fs shrinker (fs/super.c:super_cache_scan()).
This has some disadvantages and for some heavy workloads with memory
pressure it can cause some delays and stalls that make a machine
unresponsive for some periods. This happens because:
1) We can have several kswapd tasks on machines with multiple NUMA zones,
and running the extent map shrinker concurrently can cause high
contention on some spin locks, namely the spin locks that protect
the radix tree that tracks roots, the per root xarray that tracks
open inodes and the list of delayed iputs. This not only delays the
shrinker but also causes high CPU consumption and makes the task
running the shrinker monopolize a core, resulting in the symptoms
of an unresponsive system. This was noted in previous commits such as
commit ae1e766f623f ("btrfs: only run the extent map shrinker from
kswapd tasks");
2) The extent map shrinker's iteration over inodes can often be slow, even
after changing the data structure that tracks open inodes for a root
from a red black tree (up to kernel 6.10) to an xarray (kernel 6.10+).
The transition to the xarray while it made things a bit faster, it's
still somewhat slow - for example in a test scenario with 10000 inodes
that have no extent maps loaded, the extent map shrinker took between
5ms to 8ms, using a release, non-debug kernel. Iterating over the
extent maps of an inode can also be slow if have an inode with many
thousands of extent maps, since we use a red black tree to track and
search extent maps. So having the extent map shrinker run synchronously
adds extra delay for other things a kswapd task does.
So make the extent map shrinker run asynchronously as a job for the
system unbounded workqueue, just like what we do for data and metadata
space reclaim jobs.
Reviewed-by: Josef Bacik <josef@toxicpanda.com>
Signed-off-by: Filipe Manana <fdmanana@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2024-08-29 15:23:32 +01:00
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btrfs: only run the extent map shrinker from kswapd tasks
Currently the extent map shrinker can be run by any task when attempting
to allocate memory and there's enough memory pressure to trigger it.
To avoid too much latency we stop iterating over extent maps and removing
them once the task needs to reschedule. This logic was introduced in commit
b3ebb9b7e92a ("btrfs: stop extent map shrinker if reschedule is needed").
While that solved high latency problems for some use cases, it's still
not enough because with a too high number of tasks entering the extent map
shrinker code, either due to memory allocations or because they are a
kswapd task, we end up having a very high level of contention on some
spin locks, namely:
1) The fs_info->fs_roots_radix_lock spin lock, which we need to find
roots to iterate over their inodes;
2) The spin lock of the xarray used to track open inodes for a root
(struct btrfs_root::inodes) - on 6.10 kernels and below, it used to
be a red black tree and the spin lock was root->inode_lock;
3) The fs_info->delayed_iput_lock spin lock since the shrinker adds
delayed iputs (calls btrfs_add_delayed_iput()).
Instead of allowing the extent map shrinker to be run by any task, make
it run only by kswapd tasks. This still solves the problem of running
into OOM situations due to an unbounded extent map creation, which is
simple to trigger by direct IO writes, as described in the changelog
of commit 956a17d9d050 ("btrfs: add a shrinker for extent maps"), and
by a similar case when doing buffered IO on files with a very large
number of holes (keeping the file open and creating many holes, whose
extent maps are only released when the file is closed).
Reported-by: kzd <kzd@56709.net>
Link: https://bugzilla.kernel.org/show_bug.cgi?id=219121
Reported-by: Octavia Togami <octavia.togami@gmail.com>
Link: https://lore.kernel.org/linux-btrfs/CAHPNGSSt-a4ZZWrtJdVyYnJFscFjP9S7rMcvEMaNSpR556DdLA@mail.gmail.com/
Fixes: 956a17d9d050 ("btrfs: add a shrinker for extent maps")
CC: stable@vger.kernel.org # 6.10+
Tested-by: kzd <kzd@56709.net>
Tested-by: Octavia Togami <octavia.togami@gmail.com>
Signed-off-by: Filipe Manana <fdmanana@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2024-08-11 11:53:42 +01:00
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btrfs: make the extent map shrinker run asynchronously as a work queue job
Currently the extent map shrinker is run synchronously for kswapd tasks
that end up calling the fs shrinker (fs/super.c:super_cache_scan()).
This has some disadvantages and for some heavy workloads with memory
pressure it can cause some delays and stalls that make a machine
unresponsive for some periods. This happens because:
1) We can have several kswapd tasks on machines with multiple NUMA zones,
and running the extent map shrinker concurrently can cause high
contention on some spin locks, namely the spin locks that protect
the radix tree that tracks roots, the per root xarray that tracks
open inodes and the list of delayed iputs. This not only delays the
shrinker but also causes high CPU consumption and makes the task
running the shrinker monopolize a core, resulting in the symptoms
of an unresponsive system. This was noted in previous commits such as
commit ae1e766f623f ("btrfs: only run the extent map shrinker from
kswapd tasks");
2) The extent map shrinker's iteration over inodes can often be slow, even
after changing the data structure that tracks open inodes for a root
from a red black tree (up to kernel 6.10) to an xarray (kernel 6.10+).
The transition to the xarray while it made things a bit faster, it's
still somewhat slow - for example in a test scenario with 10000 inodes
that have no extent maps loaded, the extent map shrinker took between
5ms to 8ms, using a release, non-debug kernel. Iterating over the
extent maps of an inode can also be slow if have an inode with many
thousands of extent maps, since we use a red black tree to track and
search extent maps. So having the extent map shrinker run synchronously
adds extra delay for other things a kswapd task does.
So make the extent map shrinker run asynchronously as a job for the
system unbounded workqueue, just like what we do for data and metadata
space reclaim jobs.
Reviewed-by: Josef Bacik <josef@toxicpanda.com>
Signed-off-by: Filipe Manana <fdmanana@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2024-08-29 15:23:32 +01:00
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|
btrfs: add a shrinker for extent maps
Extent maps are used either to represent existing file extent items, or to
represent new extents that are going to be written and the respective file
extent items are created when the ordered extent completes.
We currently don't have any limit for how many extent maps we can have,
neither per inode nor globally. Most of the time this not too noticeable
because extent maps are removed in the following situations:
1) When evicting an inode;
2) When releasing folios (pages) through the btrfs_release_folio() address
space operation callback.
However we won't release extent maps in the folio range if the folio is
either dirty or under writeback or if the inode's i_size is less than
or equals to 16M (see try_release_extent_mapping(). This 16M i_size
constraint was added back in 2008 with commit 70dec8079d78 ("Btrfs:
extent_io and extent_state optimizations"), but there's no explanation
about why we have it or why the 16M value.
This means that for buffered IO we can reach an OOM situation due to too
many extent maps if either of the following happens:
1) There's a set of tasks constantly doing IO on many files with a size
not larger than 16M, specially if they keep the files open for very
long periods, therefore preventing inode eviction.
This requires a really high number of such files, and having many non
mergeable extent maps (due to random 4K writes for example) and a
machine with very little memory;
2) There's a set tasks constantly doing random write IO (therefore
creating many non mergeable extent maps) on files and keeping them
open for long periods of time, so inode eviction doesn't happen and
there's always a lot of dirty pages or pages under writeback,
preventing btrfs_release_folio() from releasing the respective extent
maps.
This second case was actually reported in the thread pointed by the Link
tag below, and it requires a very large file under heavy IO and a machine
with very little amount of RAM, which is probably hard to happen in
practice in a real world use case.
However when using direct IO this is not so hard to happen, because the
page cache is not used, and therefore btrfs_release_folio() is never
called. Which means extent maps are dropped only when evicting the inode,
and that means that if we have tasks that keep a file descriptor open and
keep doing IO on a very large file (or files), we can exhaust memory due
to an unbounded amount of extent maps. This is especially easy to happen
if we have a huge file with millions of small extents and their extent
maps are not mergeable (non contiguous offsets and disk locations).
This was reported in that thread with the following fio test:
$ cat test.sh
#!/bin/bash
DEV=/dev/sdj
MNT=/mnt/sdj
MOUNT_OPTIONS="-o ssd"
MKFS_OPTIONS=""
cat <<EOF > /tmp/fio-job.ini
[global]
name=fio-rand-write
filename=$MNT/fio-rand-write
rw=randwrite
bs=4K
direct=1
numjobs=16
fallocate=none
time_based
runtime=90000
[file1]
size=300G
ioengine=libaio
iodepth=16
EOF
umount $MNT &> /dev/null
mkfs.btrfs -f $MKFS_OPTIONS $DEV
mount $MOUNT_OPTIONS $DEV $MNT
fio /tmp/fio-job.ini
umount $MNT
Monitoring the btrfs_extent_map slab while running the test with:
$ watch -d -n 1 'cat /sys/kernel/slab/btrfs_extent_map/objects \
/sys/kernel/slab/btrfs_extent_map/total_objects'
Shows the number of active and total extent maps skyrocketing to tens of
millions, and on systems with a short amount of memory it's easy and quick
to get into an OOM situation, as reported in that thread.
So to avoid this issue add a shrinker that will remove extents maps, as
long as they are not pinned, and takes proper care with any concurrent
fsync to avoid missing extents (setting the full sync flag while in the
middle of a fast fsync). This shrinker is triggered through the callbacks
nr_cached_objects and free_cached_objects of struct super_operations.
The shrinker will iterate over all roots and over all inodes of each
root, and keeps track of the last scanned root and inode, so that the
next time it runs, it starts from that root and from the next inode.
This is similar to what xfs does for its inode reclaim (implements those
callbacks, and cycles through inodes by starting from where it ended
last time).
Reviewed-by: Josef Bacik <josef@toxicpanda.com>
Signed-off-by: Filipe Manana <fdmanana@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2024-04-15 17:09:26 +01:00
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btrfs: implement remove_bdev and shutdown super operation callbacks
For the ->remove_bdev() callback, btrfs will:
- Mark the target device as missing
- Go degraded if the fs can afford it
- Return error other wise
Thus falls back to the shutdown callback
For the ->shutdown callback, btrfs will:
- Set the SHUTDOWN flag
Which will reject all new incoming operations, and make all writeback
to fail.
The behavior is the same as the NOLOGFLUSH behavior.
To support the lookup from bdev to a btrfs_device,
btrfs_dev_lookup_args is enhanced to have a new @devt member.
If set, we should be able to use that @devt member to uniquely locating a
btrfs device.
I know the shutdown can be a little overkilled, if one has a RAID1
metadata and RAID0 data, in that case one can still read data with 50%
chance to got some good data.
But a filesystem returning -EIO for half of the time is not really
considered usable.
Further it can also be as bad as the only device went missing for a single
device btrfs.
So here we go safe other than sorry when handling missing device.
And the remove_bdev callback will be hidden behind experimental features
for now, the reasons are:
- There are not enough btrfs specific bdev removal test cases
The existing test cases are all removing the only device, thus only
exercises the ->shutdown() behavior.
- Not yet determined what's the expected behavior
Although the current auto-degrade behavior is no worse than the old
behavior, it may not always be what the end users want.
Before there is a concrete interface, better hide the new feature
from end users.
Reviewed-by: Johannes Thumshirn <johannes.thumshirn@wdc.com>
Reviewed-by: Anand Jain <asj@kernel.org>
Tested-by: Anand Jain <asj@kernel.org>
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2025-10-13 10:22:05 +10:30
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btrfs: zoned: show statistics about zoned filesystems in mountstats
Add statistics output to /proc/<pid>/mountstats for zoned BTRFS, similar
to the zoned statistics from XFS in mountstats.
The output for /proc/<pid>/mountstats on an example filesystem will be as
follows:
device /dev/vda mounted on /mnt with fstype btrfs
zoned statistics:
active block-groups: 7
reclaimable: 0
unused: 5
need reclaim: false
data relocation block-group: 1342177280
active zones:
start: 1073741824, wp: 268419072 used: 0, reserved: 268419072, unusable: 0
start: 1342177280, wp: 0 used: 0, reserved: 0, unusable: 0
start: 1610612736, wp: 49152 used: 16384, reserved: 16384, unusable: 16384
start: 1879048192, wp: 950272 used: 131072, reserved: 622592, unusable: 196608
start: 2147483648, wp: 212238336 used: 0, reserved: 212238336, unusable: 0
start: 2415919104, wp: 0 used: 0, reserved: 0, unusable: 0
start: 2684354560, wp: 0 used: 0, reserved: 0, unusable: 0
Reviewed-by: Naohiro Aota <naohiro.aota@wdc.com>
Reviewed-by: Filipe Manana <fdmanana@suse.com>
Signed-off-by: Johannes Thumshirn <johannes.thumshirn@wdc.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2025-12-17 14:41:37 +01:00
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2009-09-21 17:01:09 -07:00
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2009-09-21 16:00:26 -04:00
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2010-06-07 11:35:40 -04:00
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2007-03-22 12:13:20 -04:00
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2007-03-23 10:01:08 -04:00
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2009-04-02 16:46:06 -04:00
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2012-06-05 14:13:12 -04:00
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2007-04-02 10:50:19 -04:00
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2019-04-10 15:14:41 -04:00
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2007-04-19 21:01:03 -04:00
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2009-01-10 06:09:52 -08:00
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btrfs: fix fsfreeze hang caused by delayed iputs deal
When running fstests generic/068, sometimes we got below deadlock:
xfs_io D ffff8800331dbb20 0 6697 6693 0x00000080
ffff8800331dbb20 ffff88007acfc140 ffff880034d895c0 ffff8800331dc000
ffff880032d243e8 fffffffeffffffff ffff880032d24400 0000000000000001
ffff8800331dbb38 ffffffff816a9045 ffff880034d895c0 ffff8800331dbba8
Call Trace:
[<ffffffff816a9045>] schedule+0x35/0x80
[<ffffffff816abab2>] rwsem_down_read_failed+0xf2/0x140
[<ffffffff8118f5e1>] ? __filemap_fdatawrite_range+0xd1/0x100
[<ffffffff8134f978>] call_rwsem_down_read_failed+0x18/0x30
[<ffffffffa06631fc>] ? btrfs_alloc_block_rsv+0x2c/0xb0 [btrfs]
[<ffffffff810d32b5>] percpu_down_read+0x35/0x50
[<ffffffff81217dfc>] __sb_start_write+0x2c/0x40
[<ffffffffa067f5d5>] start_transaction+0x2a5/0x4d0 [btrfs]
[<ffffffffa067f857>] btrfs_join_transaction+0x17/0x20 [btrfs]
[<ffffffffa068ba34>] btrfs_evict_inode+0x3c4/0x5d0 [btrfs]
[<ffffffff81230a1a>] evict+0xba/0x1a0
[<ffffffff812316b6>] iput+0x196/0x200
[<ffffffffa06851d0>] btrfs_run_delayed_iputs+0x70/0xc0 [btrfs]
[<ffffffffa067f1d8>] btrfs_commit_transaction+0x928/0xa80 [btrfs]
[<ffffffffa0646df0>] btrfs_freeze+0x30/0x40 [btrfs]
[<ffffffff81218040>] freeze_super+0xf0/0x190
[<ffffffff81229275>] do_vfs_ioctl+0x4a5/0x5c0
[<ffffffff81003176>] ? do_audit_syscall_entry+0x66/0x70
[<ffffffff810038cf>] ? syscall_trace_enter_phase1+0x11f/0x140
[<ffffffff81229409>] SyS_ioctl+0x79/0x90
[<ffffffff81003c12>] do_syscall_64+0x62/0x110
[<ffffffff816acbe1>] entry_SYSCALL64_slow_path+0x25/0x25
>From this warning, freeze_super() already holds SB_FREEZE_FS, but
btrfs_freeze() will call btrfs_commit_transaction() again, if
btrfs_commit_transaction() finds that it has delayed iputs to handle,
it'll start_transaction(), which will try to get SB_FREEZE_FS lock
again, then deadlock occurs.
The root cause is that in btrfs, sync_filesystem(sb) does not make
sure all metadata is updated. There still maybe some codes adding
delayed iputs, see below sample race window:
CPU1 | CPU2
|-> freeze_super() |
|-> sync_filesystem(sb); |
| |-> cleaner_kthread()
| | |-> btrfs_delete_unused_bgs()
| | |-> btrfs_remove_chunk()
| | |-> btrfs_remove_block_group()
| | |-> btrfs_add_delayed_iput()
| |
|-> sb->s_writers.frozen = SB_FREEZE_FS; |
|-> sb_wait_write(sb, SB_FREEZE_FS); |
| acquire SB_FREEZE_FS lock. |
| |
|-> btrfs_freeze() |
|-> btrfs_commit_transaction() |
|-> btrfs_run_delayed_iputs() |
| will handle delayed iputs, |
| that means start_transaction() |
| will be called, which will try |
| to get SB_FREEZE_FS lock. |
To fix this issue, introduce a "int fs_frozen" to record internally whether
fs has been frozen. If fs has been frozen, we can not handle delayed iputs.
Signed-off-by: Wang Xiaoguang <wangxg.fnst@cn.fujitsu.com>
Reviewed-by: David Sterba <dsterba@suse.com>
[ add comment to btrfs_freeze ]
Signed-off-by: David Sterba <dsterba@suse.com>
Signed-off-by: Chris Mason <clm@fb.com>
2016-08-01 13:28:08 +08:00
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btrfs: add a shrinker for extent maps
Extent maps are used either to represent existing file extent items, or to
represent new extents that are going to be written and the respective file
extent items are created when the ordered extent completes.
We currently don't have any limit for how many extent maps we can have,
neither per inode nor globally. Most of the time this not too noticeable
because extent maps are removed in the following situations:
1) When evicting an inode;
2) When releasing folios (pages) through the btrfs_release_folio() address
space operation callback.
However we won't release extent maps in the folio range if the folio is
either dirty or under writeback or if the inode's i_size is less than
or equals to 16M (see try_release_extent_mapping(). This 16M i_size
constraint was added back in 2008 with commit 70dec8079d78 ("Btrfs:
extent_io and extent_state optimizations"), but there's no explanation
about why we have it or why the 16M value.
This means that for buffered IO we can reach an OOM situation due to too
many extent maps if either of the following happens:
1) There's a set of tasks constantly doing IO on many files with a size
not larger than 16M, specially if they keep the files open for very
long periods, therefore preventing inode eviction.
This requires a really high number of such files, and having many non
mergeable extent maps (due to random 4K writes for example) and a
machine with very little memory;
2) There's a set tasks constantly doing random write IO (therefore
creating many non mergeable extent maps) on files and keeping them
open for long periods of time, so inode eviction doesn't happen and
there's always a lot of dirty pages or pages under writeback,
preventing btrfs_release_folio() from releasing the respective extent
maps.
This second case was actually reported in the thread pointed by the Link
tag below, and it requires a very large file under heavy IO and a machine
with very little amount of RAM, which is probably hard to happen in
practice in a real world use case.
However when using direct IO this is not so hard to happen, because the
page cache is not used, and therefore btrfs_release_folio() is never
called. Which means extent maps are dropped only when evicting the inode,
and that means that if we have tasks that keep a file descriptor open and
keep doing IO on a very large file (or files), we can exhaust memory due
to an unbounded amount of extent maps. This is especially easy to happen
if we have a huge file with millions of small extents and their extent
maps are not mergeable (non contiguous offsets and disk locations).
This was reported in that thread with the following fio test:
$ cat test.sh
#!/bin/bash
DEV=/dev/sdj
MNT=/mnt/sdj
MOUNT_OPTIONS="-o ssd"
MKFS_OPTIONS=""
cat <<EOF > /tmp/fio-job.ini
[global]
name=fio-rand-write
filename=$MNT/fio-rand-write
rw=randwrite
bs=4K
direct=1
numjobs=16
fallocate=none
time_based
runtime=90000
[file1]
size=300G
ioengine=libaio
iodepth=16
EOF
umount $MNT &> /dev/null
mkfs.btrfs -f $MKFS_OPTIONS $DEV
mount $MOUNT_OPTIONS $DEV $MNT
fio /tmp/fio-job.ini
umount $MNT
Monitoring the btrfs_extent_map slab while running the test with:
$ watch -d -n 1 'cat /sys/kernel/slab/btrfs_extent_map/objects \
/sys/kernel/slab/btrfs_extent_map/total_objects'
Shows the number of active and total extent maps skyrocketing to tens of
millions, and on systems with a short amount of memory it's easy and quick
to get into an OOM situation, as reported in that thread.
So to avoid this issue add a shrinker that will remove extents maps, as
long as they are not pinned, and takes proper care with any concurrent
fsync to avoid missing extents (setting the full sync flag while in the
middle of a fast fsync). This shrinker is triggered through the callbacks
nr_cached_objects and free_cached_objects of struct super_operations.
The shrinker will iterate over all roots and over all inodes of each
root, and keeps track of the last scanned root and inode, so that the
next time it runs, it starts from that root and from the next inode.
This is similar to what xfs does for its inode reclaim (implements those
callbacks, and cycles through inodes by starting from where it ended
last time).
Reviewed-by: Josef Bacik <josef@toxicpanda.com>
Signed-off-by: Filipe Manana <fdmanana@suse.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2024-04-15 17:09:26 +01:00
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btrfs: zoned: show statistics about zoned filesystems in mountstats
Add statistics output to /proc/<pid>/mountstats for zoned BTRFS, similar
to the zoned statistics from XFS in mountstats.
The output for /proc/<pid>/mountstats on an example filesystem will be as
follows:
device /dev/vda mounted on /mnt with fstype btrfs
zoned statistics:
active block-groups: 7
reclaimable: 0
unused: 5
need reclaim: false
data relocation block-group: 1342177280
active zones:
start: 1073741824, wp: 268419072 used: 0, reserved: 268419072, unusable: 0
start: 1342177280, wp: 0 used: 0, reserved: 0, unusable: 0
start: 1610612736, wp: 49152 used: 16384, reserved: 16384, unusable: 16384
start: 1879048192, wp: 950272 used: 131072, reserved: 622592, unusable: 196608
start: 2147483648, wp: 212238336 used: 0, reserved: 212238336, unusable: 0
start: 2415919104, wp: 0 used: 0, reserved: 0, unusable: 0
start: 2684354560, wp: 0 used: 0, reserved: 0, unusable: 0
Reviewed-by: Naohiro Aota <naohiro.aota@wdc.com>
Reviewed-by: Filipe Manana <fdmanana@suse.com>
Signed-off-by: Johannes Thumshirn <johannes.thumshirn@wdc.com>
Reviewed-by: David Sterba <dsterba@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2025-12-17 14:41:37 +01:00
|
|
|
|
btrfs: implement remove_bdev and shutdown super operation callbacks
For the ->remove_bdev() callback, btrfs will:
- Mark the target device as missing
- Go degraded if the fs can afford it
- Return error other wise
Thus falls back to the shutdown callback
For the ->shutdown callback, btrfs will:
- Set the SHUTDOWN flag
Which will reject all new incoming operations, and make all writeback
to fail.
The behavior is the same as the NOLOGFLUSH behavior.
To support the lookup from bdev to a btrfs_device,
btrfs_dev_lookup_args is enhanced to have a new @devt member.
If set, we should be able to use that @devt member to uniquely locating a
btrfs device.
I know the shutdown can be a little overkilled, if one has a RAID1
metadata and RAID0 data, in that case one can still read data with 50%
chance to got some good data.
But a filesystem returning -EIO for half of the time is not really
considered usable.
Further it can also be as bad as the only device went missing for a single
device btrfs.
So here we go safe other than sorry when handling missing device.
And the remove_bdev callback will be hidden behind experimental features
for now, the reasons are:
- There are not enough btrfs specific bdev removal test cases
The existing test cases are all removing the only device, thus only
exercises the ->shutdown() behavior.
- Not yet determined what's the expected behavior
Although the current auto-degrade behavior is no worse than the old
behavior, it may not always be what the end users want.
Before there is a concrete interface, better hide the new feature
from end users.
Reviewed-by: Johannes Thumshirn <johannes.thumshirn@wdc.com>
Reviewed-by: Anand Jain <asj@kernel.org>
Tested-by: Anand Jain <asj@kernel.org>
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2025-10-13 10:22:05 +10:30
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2007-03-22 12:13:20 -04:00
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2008-03-24 15:02:04 -04:00
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2015-03-24 16:35:49 +01:00
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2008-03-24 15:02:04 -04:00
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2018-09-11 21:59:08 +02:00
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2008-03-24 15:02:04 -04:00
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|
llseek: automatically add .llseek fop
All file_operations should get a .llseek operation so we can make
nonseekable_open the default for future file operations without a
.llseek pointer.
The three cases that we can automatically detect are no_llseek, seq_lseek
and default_llseek. For cases where we can we can automatically prove that
the file offset is always ignored, we use noop_llseek, which maintains
the current behavior of not returning an error from a seek.
New drivers should normally not use noop_llseek but instead use no_llseek
and call nonseekable_open at open time. Existing drivers can be converted
to do the same when the maintainer knows for certain that no user code
relies on calling seek on the device file.
The generated code is often incorrectly indented and right now contains
comments that clarify for each added line why a specific variant was
chosen. In the version that gets submitted upstream, the comments will
be gone and I will manually fix the indentation, because there does not
seem to be a way to do that using coccinelle.
Some amount of new code is currently sitting in linux-next that should get
the same modifications, which I will do at the end of the merge window.
Many thanks to Julia Lawall for helping me learn to write a semantic
patch that does all this.
===== begin semantic patch =====
// This adds an llseek= method to all file operations,
// as a preparation for making no_llseek the default.
//
// The rules are
// - use no_llseek explicitly if we do nonseekable_open
// - use seq_lseek for sequential files
// - use default_llseek if we know we access f_pos
// - use noop_llseek if we know we don't access f_pos,
// but we still want to allow users to call lseek
//
@ open1 exists @
identifier nested_open;
@@
nested_open(...)
{
<+...
nonseekable_open(...)
...+>
}
@ open exists@
identifier open_f;
identifier i, f;
identifier open1.nested_open;
@@
int open_f(struct inode *i, struct file *f)
{
<+...
(
nonseekable_open(...)
|
nested_open(...)
)
...+>
}
@ read disable optional_qualifier exists @
identifier read_f;
identifier f, p, s, off;
type ssize_t, size_t, loff_t;
expression E;
identifier func;
@@
ssize_t read_f(struct file *f, char *p, size_t s, loff_t *off)
{
<+...
(
*off = E
|
*off += E
|
func(..., off, ...)
|
E = *off
)
...+>
}
@ read_no_fpos disable optional_qualifier exists @
identifier read_f;
identifier f, p, s, off;
type ssize_t, size_t, loff_t;
@@
ssize_t read_f(struct file *f, char *p, size_t s, loff_t *off)
{
... when != off
}
@ write @
identifier write_f;
identifier f, p, s, off;
type ssize_t, size_t, loff_t;
expression E;
identifier func;
@@
ssize_t write_f(struct file *f, const char *p, size_t s, loff_t *off)
{
<+...
(
*off = E
|
*off += E
|
func(..., off, ...)
|
E = *off
)
...+>
}
@ write_no_fpos @
identifier write_f;
identifier f, p, s, off;
type ssize_t, size_t, loff_t;
@@
ssize_t write_f(struct file *f, const char *p, size_t s, loff_t *off)
{
... when != off
}
@ fops0 @
identifier fops;
@@
struct file_operations fops = {
...
};
@ has_llseek depends on fops0 @
identifier fops0.fops;
identifier llseek_f;
@@
struct file_operations fops = {
...
.llseek = llseek_f,
...
};
@ has_read depends on fops0 @
identifier fops0.fops;
identifier read_f;
@@
struct file_operations fops = {
...
.read = read_f,
...
};
@ has_write depends on fops0 @
identifier fops0.fops;
identifier write_f;
@@
struct file_operations fops = {
...
.write = write_f,
...
};
@ has_open depends on fops0 @
identifier fops0.fops;
identifier open_f;
@@
struct file_operations fops = {
...
.open = open_f,
...
};
// use no_llseek if we call nonseekable_open
////////////////////////////////////////////
@ nonseekable1 depends on !has_llseek && has_open @
identifier fops0.fops;
identifier nso ~= "nonseekable_open";
@@
struct file_operations fops = {
... .open = nso, ...
+.llseek = no_llseek, /* nonseekable */
};
@ nonseekable2 depends on !has_llseek @
identifier fops0.fops;
identifier open.open_f;
@@
struct file_operations fops = {
... .open = open_f, ...
+.llseek = no_llseek, /* open uses nonseekable */
};
// use seq_lseek for sequential files
/////////////////////////////////////
@ seq depends on !has_llseek @
identifier fops0.fops;
identifier sr ~= "seq_read";
@@
struct file_operations fops = {
... .read = sr, ...
+.llseek = seq_lseek, /* we have seq_read */
};
// use default_llseek if there is a readdir
///////////////////////////////////////////
@ fops1 depends on !has_llseek && !nonseekable1 && !nonseekable2 && !seq @
identifier fops0.fops;
identifier readdir_e;
@@
// any other fop is used that changes pos
struct file_operations fops = {
... .readdir = readdir_e, ...
+.llseek = default_llseek, /* readdir is present */
};
// use default_llseek if at least one of read/write touches f_pos
/////////////////////////////////////////////////////////////////
@ fops2 depends on !fops1 && !has_llseek && !nonseekable1 && !nonseekable2 && !seq @
identifier fops0.fops;
identifier read.read_f;
@@
// read fops use offset
struct file_operations fops = {
... .read = read_f, ...
+.llseek = default_llseek, /* read accesses f_pos */
};
@ fops3 depends on !fops1 && !fops2 && !has_llseek && !nonseekable1 && !nonseekable2 && !seq @
identifier fops0.fops;
identifier write.write_f;
@@
// write fops use offset
struct file_operations fops = {
... .write = write_f, ...
+ .llseek = default_llseek, /* write accesses f_pos */
};
// Use noop_llseek if neither read nor write accesses f_pos
///////////////////////////////////////////////////////////
@ fops4 depends on !fops1 && !fops2 && !fops3 && !has_llseek && !nonseekable1 && !nonseekable2 && !seq @
identifier fops0.fops;
identifier read_no_fpos.read_f;
identifier write_no_fpos.write_f;
@@
// write fops use offset
struct file_operations fops = {
...
.write = write_f,
.read = read_f,
...
+.llseek = noop_llseek, /* read and write both use no f_pos */
};
@ depends on has_write && !has_read && !fops1 && !fops2 && !has_llseek && !nonseekable1 && !nonseekable2 && !seq @
identifier fops0.fops;
identifier write_no_fpos.write_f;
@@
struct file_operations fops = {
... .write = write_f, ...
+.llseek = noop_llseek, /* write uses no f_pos */
};
@ depends on has_read && !has_write && !fops1 && !fops2 && !has_llseek && !nonseekable1 && !nonseekable2 && !seq @
identifier fops0.fops;
identifier read_no_fpos.read_f;
@@
struct file_operations fops = {
... .read = read_f, ...
+.llseek = noop_llseek, /* read uses no f_pos */
};
@ depends on !has_read && !has_write && !fops1 && !fops2 && !has_llseek && !nonseekable1 && !nonseekable2 && !seq @
identifier fops0.fops;
@@
struct file_operations fops = {
...
+.llseek = noop_llseek, /* no read or write fn */
};
===== End semantic patch =====
Signed-off-by: Arnd Bergmann <arnd@arndb.de>
Cc: Julia Lawall <julia@diku.dk>
Cc: Christoph Hellwig <hch@infradead.org>
2010-08-15 18:52:59 +02:00
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2008-03-24 15:02:04 -04:00
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driver core: add devname module aliases to allow module on-demand auto-loading
This adds:
alias: devname:<name>
to some common kernel modules, which will allow the on-demand loading
of the kernel module when the device node is accessed.
Ideally all these modules would be compiled-in, but distros seems too
much in love with their modularization that we need to cover the common
cases with this new facility. It will allow us to remove a bunch of pretty
useless init scripts and modprobes from init scripts.
The static device node aliases will be carried in the module itself. The
program depmod will extract this information to a file in the module directory:
$ cat /lib/modules/2.6.34-00650-g537b60d-dirty/modules.devname
# Device nodes to trigger on-demand module loading.
microcode cpu/microcode c10:184
fuse fuse c10:229
ppp_generic ppp c108:0
tun net/tun c10:200
dm_mod mapper/control c10:235
Udev will pick up the depmod created file on startup and create all the
static device nodes which the kernel modules specify, so that these modules
get automatically loaded when the device node is accessed:
$ /sbin/udevd --debug
...
static_dev_create_from_modules: mknod '/dev/cpu/microcode' c10:184
static_dev_create_from_modules: mknod '/dev/fuse' c10:229
static_dev_create_from_modules: mknod '/dev/ppp' c108:0
static_dev_create_from_modules: mknod '/dev/net/tun' c10:200
static_dev_create_from_modules: mknod '/dev/mapper/control' c10:235
udev_rules_apply_static_dev_perms: chmod '/dev/net/tun' 0666
udev_rules_apply_static_dev_perms: chmod '/dev/fuse' 0666
A few device nodes are switched to statically allocated numbers, to allow
the static nodes to work. This might also useful for systems which still run
a plain static /dev, which is completely unsafe to use with any dynamic minor
numbers.
Note:
The devname aliases must be limited to the *common* and *single*instance*
device nodes, like the misc devices, and never be used for conceptually limited
systems like the loop devices, which should rather get fixed properly and get a
control node for losetup to talk to, instead of creating a random number of
device nodes in advance, regardless if they are ever used.
This facility is to hide the mess distros are creating with too modualized
kernels, and just to hide that these modules are not compiled-in, and not to
paper-over broken concepts. Thanks! :)
Cc: Greg Kroah-Hartman <gregkh@suse.de>
Cc: David S. Miller <davem@davemloft.net>
Cc: Miklos Szeredi <miklos@szeredi.hu>
Cc: Chris Mason <chris.mason@oracle.com>
Cc: Alasdair G Kergon <agk@redhat.com>
Cc: Tigran Aivazian <tigran@aivazian.fsnet.co.uk>
Cc: Ian Kent <raven@themaw.net>
Signed-Off-By: Kay Sievers <kay.sievers@vrfy.org>
Signed-off-by: Greg Kroah-Hartman <gregkh@suse.de>
2010-05-20 18:07:20 +02:00
|
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2008-03-24 15:02:04 -04:00
|
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|
driver core: add devname module aliases to allow module on-demand auto-loading
This adds:
alias: devname:<name>
to some common kernel modules, which will allow the on-demand loading
of the kernel module when the device node is accessed.
Ideally all these modules would be compiled-in, but distros seems too
much in love with their modularization that we need to cover the common
cases with this new facility. It will allow us to remove a bunch of pretty
useless init scripts and modprobes from init scripts.
The static device node aliases will be carried in the module itself. The
program depmod will extract this information to a file in the module directory:
$ cat /lib/modules/2.6.34-00650-g537b60d-dirty/modules.devname
# Device nodes to trigger on-demand module loading.
microcode cpu/microcode c10:184
fuse fuse c10:229
ppp_generic ppp c108:0
tun net/tun c10:200
dm_mod mapper/control c10:235
Udev will pick up the depmod created file on startup and create all the
static device nodes which the kernel modules specify, so that these modules
get automatically loaded when the device node is accessed:
$ /sbin/udevd --debug
...
static_dev_create_from_modules: mknod '/dev/cpu/microcode' c10:184
static_dev_create_from_modules: mknod '/dev/fuse' c10:229
static_dev_create_from_modules: mknod '/dev/ppp' c108:0
static_dev_create_from_modules: mknod '/dev/net/tun' c10:200
static_dev_create_from_modules: mknod '/dev/mapper/control' c10:235
udev_rules_apply_static_dev_perms: chmod '/dev/net/tun' 0666
udev_rules_apply_static_dev_perms: chmod '/dev/fuse' 0666
A few device nodes are switched to statically allocated numbers, to allow
the static nodes to work. This might also useful for systems which still run
a plain static /dev, which is completely unsafe to use with any dynamic minor
numbers.
Note:
The devname aliases must be limited to the *common* and *single*instance*
device nodes, like the misc devices, and never be used for conceptually limited
systems like the loop devices, which should rather get fixed properly and get a
control node for losetup to talk to, instead of creating a random number of
device nodes in advance, regardless if they are ever used.
This facility is to hide the mess distros are creating with too modualized
kernels, and just to hide that these modules are not compiled-in, and not to
paper-over broken concepts. Thanks! :)
Cc: Greg Kroah-Hartman <gregkh@suse.de>
Cc: David S. Miller <davem@davemloft.net>
Cc: Miklos Szeredi <miklos@szeredi.hu>
Cc: Chris Mason <chris.mason@oracle.com>
Cc: Alasdair G Kergon <agk@redhat.com>
Cc: Tigran Aivazian <tigran@aivazian.fsnet.co.uk>
Cc: Ian Kent <raven@themaw.net>
Signed-Off-By: Kay Sievers <kay.sievers@vrfy.org>
Signed-off-by: Greg Kroah-Hartman <gregkh@suse.de>
2010-05-20 18:07:20 +02:00
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2017-11-02 17:21:50 -06:00
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2008-03-24 15:02:04 -04:00
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2018-02-19 17:24:18 +01:00
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2008-03-24 15:02:04 -04:00
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2015-07-30 15:59:57 -07:00
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2008-03-24 15:02:04 -04:00
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btrfs: make module init/exit match their sequence
[BACKGROUND]
In theory init_btrfs_fs() and exit_btrfs_fs() should match their
sequence, thus normally they should look like this:
init_btrfs_fs() | exit_btrfs_fs()
----------------------+------------------------
init_A(); |
init_B(); |
init_C(); |
| exit_C();
| exit_B();
| exit_A();
So is for the error path of init_btrfs_fs().
But it's not the case, some exit functions don't match their init
functions sequence in init_btrfs_fs().
Furthermore in init_btrfs_fs(), we need to have a new error label for
each new init function we added. This is not really expandable,
especially recently we may add several new functions to init_btrfs_fs().
[ENHANCEMENT]
The patch will introduce the following things to enhance the situation:
- struct init_sequence
Just a wrapper of init and exit function pointers.
The init function must use int type as return value, thus some init
functions need to be updated to return 0.
The exit function can be NULL, as there are some init sequence just
outputting a message.
- struct mod_init_seq[] array
This is a const array, recording all the initialization we need to do
in init_btrfs_fs(), and the order follows the old init_btrfs_fs().
- bool mod_init_result[] array
This is a bool array, recording if we have initialized one entry in
mod_init_seq[].
The reason to split mod_init_seq[] and mod_init_result[] is to avoid
section mismatch in reference.
All init function are in .init.text, but if mod_init_seq[] records
the @initialized member it can no longer be const, thus will be put
into .data section, and cause modpost warning.
For init_btrfs_fs() we just call all init functions in their order in
mod_init_seq[] array, and after each call, setting corresponding
mod_init_result[] to true.
For exit_btrfs_fs() and error handling path of init_btrfs_fs(), we just
iterate mod_init_seq[] in reverse order, and skip all uninitialized
entry.
With this patch, init_btrfs_fs()/exit_btrfs_fs() will be much easier to
expand and will always follow the strict order.
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2022-10-12 17:22:35 +08:00
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|
2013-04-30 16:51:59 +00:00
|
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2018-06-20 10:03:33 -07:00
|
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2025-01-02 02:06:37 +08:00
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2013-04-30 16:51:59 +00:00
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2013-09-03 15:25:27 +02:00
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2020-11-10 20:26:07 +09:00
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2021-07-28 18:10:39 +02:00
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2013-04-30 16:51:59 +00:00
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2018-06-20 10:03:33 -07:00
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2025-01-02 02:06:39 +08:00
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2023-04-05 07:49:04 +02:00
|
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|
2025-01-02 02:06:39 +08:00
|
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|
|
btrfs: make module init/exit match their sequence
[BACKGROUND]
In theory init_btrfs_fs() and exit_btrfs_fs() should match their
sequence, thus normally they should look like this:
init_btrfs_fs() | exit_btrfs_fs()
----------------------+------------------------
init_A(); |
init_B(); |
init_C(); |
| exit_C();
| exit_B();
| exit_A();
So is for the error path of init_btrfs_fs().
But it's not the case, some exit functions don't match their init
functions sequence in init_btrfs_fs().
Furthermore in init_btrfs_fs(), we need to have a new error label for
each new init function we added. This is not really expandable,
especially recently we may add several new functions to init_btrfs_fs().
[ENHANCEMENT]
The patch will introduce the following things to enhance the situation:
- struct init_sequence
Just a wrapper of init and exit function pointers.
The init function must use int type as return value, thus some init
functions need to be updated to return 0.
The exit function can be NULL, as there are some init sequence just
outputting a message.
- struct mod_init_seq[] array
This is a const array, recording all the initialization we need to do
in init_btrfs_fs(), and the order follows the old init_btrfs_fs().
- bool mod_init_result[] array
This is a bool array, recording if we have initialized one entry in
mod_init_seq[].
The reason to split mod_init_seq[] and mod_init_result[] is to avoid
section mismatch in reference.
All init function are in .init.text, but if mod_init_seq[] records
the @initialized member it can no longer be const, thus will be put
into .data section, and cause modpost warning.
For init_btrfs_fs() we just call all init functions in their order in
mod_init_seq[] array, and after each call, setting corresponding
mod_init_result[] to true.
For exit_btrfs_fs() and error handling path of init_btrfs_fs(), we just
iterate mod_init_seq[] in reverse order, and skip all uninitialized
entry.
With this patch, init_btrfs_fs()/exit_btrfs_fs() will be much easier to
expand and will always follow the strict order.
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2022-10-12 17:22:35 +08:00
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2024-06-25 15:52:28 +01:00
|
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btrfs: make module init/exit match their sequence
[BACKGROUND]
In theory init_btrfs_fs() and exit_btrfs_fs() should match their
sequence, thus normally they should look like this:
init_btrfs_fs() | exit_btrfs_fs()
----------------------+------------------------
init_A(); |
init_B(); |
init_C(); |
| exit_C();
| exit_B();
| exit_A();
So is for the error path of init_btrfs_fs().
But it's not the case, some exit functions don't match their init
functions sequence in init_btrfs_fs().
Furthermore in init_btrfs_fs(), we need to have a new error label for
each new init function we added. This is not really expandable,
especially recently we may add several new functions to init_btrfs_fs().
[ENHANCEMENT]
The patch will introduce the following things to enhance the situation:
- struct init_sequence
Just a wrapper of init and exit function pointers.
The init function must use int type as return value, thus some init
functions need to be updated to return 0.
The exit function can be NULL, as there are some init sequence just
outputting a message.
- struct mod_init_seq[] array
This is a const array, recording all the initialization we need to do
in init_btrfs_fs(), and the order follows the old init_btrfs_fs().
- bool mod_init_result[] array
This is a bool array, recording if we have initialized one entry in
mod_init_seq[].
The reason to split mod_init_seq[] and mod_init_result[] is to avoid
section mismatch in reference.
All init function are in .init.text, but if mod_init_seq[] records
the @initialized member it can no longer be const, thus will be put
into .data section, and cause modpost warning.
For init_btrfs_fs() we just call all init functions in their order in
mod_init_seq[] array, and after each call, setting corresponding
mod_init_result[] to true.
For exit_btrfs_fs() and error handling path of init_btrfs_fs(), we just
iterate mod_init_seq[] in reverse order, and skip all uninitialized
entry.
With this patch, init_btrfs_fs()/exit_btrfs_fs() will be much easier to
expand and will always follow the strict order.
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2022-10-12 17:22:35 +08:00
|
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2025-04-04 16:45:12 +01:00
|
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|
|
|
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|
|
|
btrfs: make module init/exit match their sequence
[BACKGROUND]
In theory init_btrfs_fs() and exit_btrfs_fs() should match their
sequence, thus normally they should look like this:
init_btrfs_fs() | exit_btrfs_fs()
----------------------+------------------------
init_A(); |
init_B(); |
init_C(); |
| exit_C();
| exit_B();
| exit_A();
So is for the error path of init_btrfs_fs().
But it's not the case, some exit functions don't match their init
functions sequence in init_btrfs_fs().
Furthermore in init_btrfs_fs(), we need to have a new error label for
each new init function we added. This is not really expandable,
especially recently we may add several new functions to init_btrfs_fs().
[ENHANCEMENT]
The patch will introduce the following things to enhance the situation:
- struct init_sequence
Just a wrapper of init and exit function pointers.
The init function must use int type as return value, thus some init
functions need to be updated to return 0.
The exit function can be NULL, as there are some init sequence just
outputting a message.
- struct mod_init_seq[] array
This is a const array, recording all the initialization we need to do
in init_btrfs_fs(), and the order follows the old init_btrfs_fs().
- bool mod_init_result[] array
This is a bool array, recording if we have initialized one entry in
mod_init_seq[].
The reason to split mod_init_seq[] and mod_init_result[] is to avoid
section mismatch in reference.
All init function are in .init.text, but if mod_init_seq[] records
the @initialized member it can no longer be const, thus will be put
into .data section, and cause modpost warning.
For init_btrfs_fs() we just call all init functions in their order in
mod_init_seq[] array, and after each call, setting corresponding
mod_init_result[] to true.
For exit_btrfs_fs() and error handling path of init_btrfs_fs(), we just
iterate mod_init_seq[] in reverse order, and skip all uninitialized
entry.
With this patch, init_btrfs_fs()/exit_btrfs_fs() will be much easier to
expand and will always follow the strict order.
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2022-10-12 17:22:35 +08:00
|
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|
2025-04-08 17:31:16 +01:00
|
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|
|
|
|
|
|
2025-01-02 02:06:38 +08:00
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
btrfs: make module init/exit match their sequence
[BACKGROUND]
In theory init_btrfs_fs() and exit_btrfs_fs() should match their
sequence, thus normally they should look like this:
init_btrfs_fs() | exit_btrfs_fs()
----------------------+------------------------
init_A(); |
init_B(); |
init_C(); |
| exit_C();
| exit_B();
| exit_A();
So is for the error path of init_btrfs_fs().
But it's not the case, some exit functions don't match their init
functions sequence in init_btrfs_fs().
Furthermore in init_btrfs_fs(), we need to have a new error label for
each new init function we added. This is not really expandable,
especially recently we may add several new functions to init_btrfs_fs().
[ENHANCEMENT]
The patch will introduce the following things to enhance the situation:
- struct init_sequence
Just a wrapper of init and exit function pointers.
The init function must use int type as return value, thus some init
functions need to be updated to return 0.
The exit function can be NULL, as there are some init sequence just
outputting a message.
- struct mod_init_seq[] array
This is a const array, recording all the initialization we need to do
in init_btrfs_fs(), and the order follows the old init_btrfs_fs().
- bool mod_init_result[] array
This is a bool array, recording if we have initialized one entry in
mod_init_seq[].
The reason to split mod_init_seq[] and mod_init_result[] is to avoid
section mismatch in reference.
All init function are in .init.text, but if mod_init_seq[] records
the @initialized member it can no longer be const, thus will be put
into .data section, and cause modpost warning.
For init_btrfs_fs() we just call all init functions in their order in
mod_init_seq[] array, and after each call, setting corresponding
mod_init_result[] to true.
For exit_btrfs_fs() and error handling path of init_btrfs_fs(), we just
iterate mod_init_seq[] in reverse order, and skip all uninitialized
entry.
With this patch, init_btrfs_fs()/exit_btrfs_fs() will be much easier to
expand and will always follow the strict order.
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2022-10-12 17:22:35 +08:00
|
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2022-10-19 21:51:42 +05:30
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btrfs: make module init/exit match their sequence
[BACKGROUND]
In theory init_btrfs_fs() and exit_btrfs_fs() should match their
sequence, thus normally they should look like this:
init_btrfs_fs() | exit_btrfs_fs()
----------------------+------------------------
init_A(); |
init_B(); |
init_C(); |
| exit_C();
| exit_B();
| exit_A();
So is for the error path of init_btrfs_fs().
But it's not the case, some exit functions don't match their init
functions sequence in init_btrfs_fs().
Furthermore in init_btrfs_fs(), we need to have a new error label for
each new init function we added. This is not really expandable,
especially recently we may add several new functions to init_btrfs_fs().
[ENHANCEMENT]
The patch will introduce the following things to enhance the situation:
- struct init_sequence
Just a wrapper of init and exit function pointers.
The init function must use int type as return value, thus some init
functions need to be updated to return 0.
The exit function can be NULL, as there are some init sequence just
outputting a message.
- struct mod_init_seq[] array
This is a const array, recording all the initialization we need to do
in init_btrfs_fs(), and the order follows the old init_btrfs_fs().
- bool mod_init_result[] array
This is a bool array, recording if we have initialized one entry in
mod_init_seq[].
The reason to split mod_init_seq[] and mod_init_result[] is to avoid
section mismatch in reference.
All init function are in .init.text, but if mod_init_seq[] records
the @initialized member it can no longer be const, thus will be put
into .data section, and cause modpost warning.
For init_btrfs_fs() we just call all init functions in their order in
mod_init_seq[] array, and after each call, setting corresponding
mod_init_result[] to true.
For exit_btrfs_fs() and error handling path of init_btrfs_fs(), we just
iterate mod_init_seq[] in reverse order, and skip all uninitialized
entry.
With this patch, init_btrfs_fs()/exit_btrfs_fs() will be much easier to
expand and will always follow the strict order.
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2022-10-12 17:22:35 +08:00
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2013-04-30 16:51:59 +00:00
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2022-10-19 21:51:42 +05:30
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2022-12-13 10:42:26 +00:00
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2022-10-19 21:51:42 +05:30
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2007-03-21 11:12:56 -04:00
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btrfs: make module init/exit match their sequence
[BACKGROUND]
In theory init_btrfs_fs() and exit_btrfs_fs() should match their
sequence, thus normally they should look like this:
init_btrfs_fs() | exit_btrfs_fs()
----------------------+------------------------
init_A(); |
init_B(); |
init_C(); |
| exit_C();
| exit_B();
| exit_A();
So is for the error path of init_btrfs_fs().
But it's not the case, some exit functions don't match their init
functions sequence in init_btrfs_fs().
Furthermore in init_btrfs_fs(), we need to have a new error label for
each new init function we added. This is not really expandable,
especially recently we may add several new functions to init_btrfs_fs().
[ENHANCEMENT]
The patch will introduce the following things to enhance the situation:
- struct init_sequence
Just a wrapper of init and exit function pointers.
The init function must use int type as return value, thus some init
functions need to be updated to return 0.
The exit function can be NULL, as there are some init sequence just
outputting a message.
- struct mod_init_seq[] array
This is a const array, recording all the initialization we need to do
in init_btrfs_fs(), and the order follows the old init_btrfs_fs().
- bool mod_init_result[] array
This is a bool array, recording if we have initialized one entry in
mod_init_seq[].
The reason to split mod_init_seq[] and mod_init_result[] is to avoid
section mismatch in reference.
All init function are in .init.text, but if mod_init_seq[] records
the @initialized member it can no longer be const, thus will be put
into .data section, and cause modpost warning.
For init_btrfs_fs() we just call all init functions in their order in
mod_init_seq[] array, and after each call, setting corresponding
mod_init_result[] to true.
For exit_btrfs_fs() and error handling path of init_btrfs_fs(), we just
iterate mod_init_seq[] in reverse order, and skip all uninitialized
entry.
With this patch, init_btrfs_fs()/exit_btrfs_fs() will be much easier to
expand and will always follow the strict order.
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2022-10-12 17:22:35 +08:00
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2013-03-15 09:47:08 -04:00
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btrfs: make module init/exit match their sequence
[BACKGROUND]
In theory init_btrfs_fs() and exit_btrfs_fs() should match their
sequence, thus normally they should look like this:
init_btrfs_fs() | exit_btrfs_fs()
----------------------+------------------------
init_A(); |
init_B(); |
init_C(); |
| exit_C();
| exit_B();
| exit_A();
So is for the error path of init_btrfs_fs().
But it's not the case, some exit functions don't match their init
functions sequence in init_btrfs_fs().
Furthermore in init_btrfs_fs(), we need to have a new error label for
each new init function we added. This is not really expandable,
especially recently we may add several new functions to init_btrfs_fs().
[ENHANCEMENT]
The patch will introduce the following things to enhance the situation:
- struct init_sequence
Just a wrapper of init and exit function pointers.
The init function must use int type as return value, thus some init
functions need to be updated to return 0.
The exit function can be NULL, as there are some init sequence just
outputting a message.
- struct mod_init_seq[] array
This is a const array, recording all the initialization we need to do
in init_btrfs_fs(), and the order follows the old init_btrfs_fs().
- bool mod_init_result[] array
This is a bool array, recording if we have initialized one entry in
mod_init_seq[].
The reason to split mod_init_seq[] and mod_init_result[] is to avoid
section mismatch in reference.
All init function are in .init.text, but if mod_init_seq[] records
the @initialized member it can no longer be const, thus will be put
into .data section, and cause modpost warning.
For init_btrfs_fs() we just call all init functions in their order in
mod_init_seq[] array, and after each call, setting corresponding
mod_init_result[] to true.
For exit_btrfs_fs() and error handling path of init_btrfs_fs(), we just
iterate mod_init_seq[] in reverse order, and skip all uninitialized
entry.
With this patch, init_btrfs_fs()/exit_btrfs_fs() will be much easier to
expand and will always follow the strict order.
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2022-10-12 17:22:35 +08:00
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2022-10-19 21:51:42 +05:30
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|
|
btrfs: make module init/exit match their sequence
[BACKGROUND]
In theory init_btrfs_fs() and exit_btrfs_fs() should match their
sequence, thus normally they should look like this:
init_btrfs_fs() | exit_btrfs_fs()
----------------------+------------------------
init_A(); |
init_B(); |
init_C(); |
| exit_C();
| exit_B();
| exit_A();
So is for the error path of init_btrfs_fs().
But it's not the case, some exit functions don't match their init
functions sequence in init_btrfs_fs().
Furthermore in init_btrfs_fs(), we need to have a new error label for
each new init function we added. This is not really expandable,
especially recently we may add several new functions to init_btrfs_fs().
[ENHANCEMENT]
The patch will introduce the following things to enhance the situation:
- struct init_sequence
Just a wrapper of init and exit function pointers.
The init function must use int type as return value, thus some init
functions need to be updated to return 0.
The exit function can be NULL, as there are some init sequence just
outputting a message.
- struct mod_init_seq[] array
This is a const array, recording all the initialization we need to do
in init_btrfs_fs(), and the order follows the old init_btrfs_fs().
- bool mod_init_result[] array
This is a bool array, recording if we have initialized one entry in
mod_init_seq[].
The reason to split mod_init_seq[] and mod_init_result[] is to avoid
section mismatch in reference.
All init function are in .init.text, but if mod_init_seq[] records
the @initialized member it can no longer be const, thus will be put
into .data section, and cause modpost warning.
For init_btrfs_fs() we just call all init functions in their order in
mod_init_seq[] array, and after each call, setting corresponding
mod_init_result[] to true.
For exit_btrfs_fs() and error handling path of init_btrfs_fs(), we just
iterate mod_init_seq[] in reverse order, and skip all uninitialized
entry.
With this patch, init_btrfs_fs()/exit_btrfs_fs() will be much easier to
expand and will always follow the strict order.
Signed-off-by: Qu Wenruo <wqu@suse.com>
Signed-off-by: David Sterba <dsterba@suse.com>
2022-10-12 17:22:35 +08:00
|
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|
2007-11-19 10:22:33 -05:00
|
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|
2007-03-21 11:12:56 -04:00
|
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|
2014-02-01 21:27:56 +00:00
|
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2007-03-21 11:12:56 -04:00
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2024-05-27 10:56:59 -07:00
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2007-03-21 11:12:56 -04:00
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