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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2007-03-22 12:13:20 -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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2007-03-26 16:00:06 -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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2011-05-26 10:01:56 -06:00
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2011-11-30 10:45:38 -05:00
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2013-01-29 06:04:50 +00:00
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2008-11-20 10:22:27 -05:00
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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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2007-05-24 13:35:57 -04:00
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2007-11-16 11:45:54 -05:00
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2008-03-24 15:02:07 -04:00
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2008-07-23 12:12:13 -04:00
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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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2012-06-05 14:13:12 -04:00
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2012-11-06 13:15:27 +01:00
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2007-03-21 11:12:56 -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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2009-09-21 17:01:09 -07:00
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2011-07-25 15:55:42 -04:00
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2007-03-22 12:13:20 -04:00
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2013-01-31 00:54:54 +00:00
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2011-01-06 19:30:25 +08:00
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2011-10-03 23:22:31 -04:00
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2011-01-06 19:30:25 +08:00
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2013-01-29 10:14:48 +00:00
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2011-01-06 19:30:25 +08:00
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2013-01-29 10:14:48 +00:00
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2011-01-06 19:30:25 +08:00
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2012-11-05 17:11:06 +01:00
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2012-03-01 17:24:58 +01:00
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2011-01-06 19:30:25 +08:00
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2012-07-30 14:40:13 -07:00
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2011-01-06 19:30:25 +08:00
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2012-03-01 14:57:30 +01:00
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2011-01-06 19:30:25 +08:00
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2012-03-01 14:57:30 +01:00
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2011-01-06 19:30:25 +08:00
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2013-01-31 00:54:54 +00:00
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2012-03-01 14:57:30 +01:00
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2013-01-31 00:54:58 +00:00
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2011-01-06 19:30:25 +08:00
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2012-03-01 14:57:30 +01:00
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2013-01-31 00:54:58 +00:00
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2012-03-01 14:57:30 +01:00
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2012-07-30 14:40:13 -07:00
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2012-03-01 14:57:30 +01:00
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2012-07-30 14:40:13 -07:00
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2012-03-01 14:57:30 +01:00
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2012-07-30 14:40:13 -07:00
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2012-03-01 14:57:30 +01:00
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2012-07-30 14:40:13 -07:00
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2012-03-01 14:57:30 +01:00
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2012-07-30 14:40:13 -07:00
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2011-01-06 19:30:25 +08:00
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2012-07-30 14:40:13 -07:00
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2012-03-01 17:24:58 +01:00
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2012-05-07 06:35:38 -06:00
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2012-03-01 17:24:58 +01:00
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2012-09-06 04:04:44 -06:00
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2013-01-31 00:54:54 +00:00
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2012-09-06 04:04:44 -06:00
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2011-01-06 19:30:25 +08:00
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2012-03-01 17:24:58 +01:00
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2013-01-15 06:27:25 +00:00
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2012-03-01 17:24:58 +01:00
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2011-10-03 23:22:31 -04:00
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2011-01-06 19:30:25 +08:00
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2013-01-31 00:54:54 +00:00
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2013-01-31 00:54:55 +00:00
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2011-10-03 23:22:31 -04:00
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2011-01-06 19:30:25 +08:00
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2011-10-03 23:22:31 -04:00
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2011-01-06 19:30:25 +08:00
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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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2011-11-17 15:40:49 -05:00
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2011-11-17 01:22:46 -05:00
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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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2007-08-29 09:11:44 -04:00
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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
|
|
|
|
2010-03-19 18:07:23 +00:00
|
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|
2010-12-17 14:21:50 +08:00
|
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|
2011-02-16 13:10:41 -05:00
|
|
|
|
2012-01-16 22:04:48 +02:00
|
|
|
|
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|
|
2011-11-09 13:44:05 +01:00
|
|
|
|
2011-10-03 23:22:31 -04:00
|
|
|
|
2012-01-16 22:04:48 +02:00
|
|
|
|
2007-08-29 09:11:44 -04:00
|
|
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|
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|
|
|
|
2008-05-13 13:46:40 -04:00
|
|
|
|
2007-08-29 09:11:44 -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
|
|
|
|
2008-06-10 10:40:46 -04:00
|
|
|
|
2007-12-14 15:30:32 -05:00
|
|
|
|
2007-12-17 20:14:01 -05:00
|
|
|
|
2008-01-09 09:23:21 -05:00
|
|
|
|
2008-01-29 16:03:38 -05:00
|
|
|
|
2008-01-02 10:01:11 -05:00
|
|
|
|
2008-06-11 21:47:56 -04:00
|
|
|
|
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
|
|
|
|
2010-12-17 14:21:50 +08:00
|
|
|
|
2010-01-28 16:18:15 -05:00
|
|
|
|
2010-12-17 14:21:50 +08:00
|
|
|
|
2008-01-18 10:54:22 -05:00
|
|
|
|
2009-06-09 20:28:34 -04:00
|
|
|
|
2009-06-09 16:42:22 -04:00
|
|
|
|
2008-07-24 12:16:36 -04:00
|
|
|
|
2009-04-02 16:49:40 -04:00
|
|
|
|
2009-04-02 16:59:01 -04:00
|
|
|
|
2009-04-21 17:40:57 -04:00
|
|
|
|
2009-10-14 09:24:59 -04:00
|
|
|
|
2010-06-21 14:48:16 -04:00
|
|
|
|
2010-09-21 14:21:34 -04:00
|
|
|
|
2010-10-29 15:46:43 -04:00
|
|
|
|
2011-02-16 13:10:41 -05:00
|
|
|
|
2011-04-06 07:33:51 +00:00
|
|
|
|
2011-05-24 15:35:30 -04:00
|
|
|
|
2011-06-03 09:36:29 -04:00
|
|
|
|
2011-11-11 10:14:57 -05:00
|
|
|
|
2011-11-03 15:17:42 -04:00
|
|
|
|
2012-01-16 22:04:48 +02:00
|
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|
|
2011-11-09 13:44:05 +01:00
|
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|
2011-10-03 23:22:31 -04:00
|
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|
2008-07-24 12:16:36 -04:00
|
|
|
|
2007-08-29 09:11:44 -04:00
|
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2008-06-10 10:40:29 -04:00
|
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|
2012-03-01 17:24:58 +01:00
|
|
|
|
2008-06-10 10:40:29 -04:00
|
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|
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|
|
2007-08-29 09:11:44 -04:00
|
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2008-06-10 10:40:29 -04:00
|
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2007-08-29 09:11:44 -04:00
|
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2011-10-03 14:07:49 -04:00
|
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2008-06-11 21:47:56 -04:00
|
|
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|
2009-11-07 06:19:16 +00:00
|
|
|
|
2010-12-17 14:21:50 +08:00
|
|
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|
|
|
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|
|
2007-12-14 15:30:32 -05:00
|
|
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|
2011-04-13 15:41:04 +02:00
|
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|
|
2011-10-03 14:07:49 -04:00
|
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2007-08-29 09:11:44 -04:00
|
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2011-10-03 14:07:49 -04:00
|
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2007-08-29 09:11:44 -04:00
|
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2007-12-17 20:14:01 -05:00
|
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2010-02-25 20:38:35 +00:00
|
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|
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2007-12-17 20:14:01 -05:00
|
|
|
|
2008-06-10 10:40:29 -04:00
|
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2007-08-29 09:11:44 -04:00
|
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2008-05-13 13:46:40 -04:00
|
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2008-06-10 10:40:29 -04:00
|
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|
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|
|
|
|
2008-05-13 13:46:40 -04:00
|
|
|
|
2007-08-29 09:11:44 -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
|
|
|
|
2011-04-06 07:33:51 +00:00
|
|
|
|
2008-06-10 10:40:46 -04:00
|
|
|
|
2008-06-10 10:40:29 -04:00
|
|
|
|
2008-06-10 10:40:46 -04:00
|
|
|
|
2008-06-10 10:40:29 -04:00
|
|
|
|
|
|
|
|
|
2007-12-14 15:30:32 -05:00
|
|
|
|
|
|
|
|
|
2009-06-11 09:30:13 -04:00
|
|
|
|
2008-06-10 10:40:29 -04:00
|
|
|
|
2007-12-17 20:14:01 -05:00
|
|
|
|
|
|
|
|
|
2012-09-20 08:42:11 -06:00
|
|
|
|
|
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2008-06-10 10:40:29 -04:00
|
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|
|
|
|
|
|
2007-08-29 09:11:44 -04:00
|
|
|
|
2010-01-28 16:18:15 -05:00
|
|
|
|
2010-12-17 14:21:50 +08:00
|
|
|
|
|
|
|
|
|
2013-01-31 00:54:56 +00:00
|
|
|
|
2010-12-17 14:21:50 +08:00
|
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|
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2012-04-16 15:27:51 +02:00
|
|
|
|
2012-09-20 08:42:11 -06:00
|
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|
|
|
|
|
|
|
2010-10-25 15:12:26 +08:00
|
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2012-04-16 15:27:51 +02:00
|
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2012-09-20 08:42:11 -06:00
|
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|
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|
|
2012-07-24 11:58:43 -06:00
|
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2012-04-16 15:27:51 +02:00
|
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2010-12-17 14:21:50 +08:00
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2010-01-28 16:18:15 -05:00
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2008-01-18 10:54:22 -05:00
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2008-06-10 10:40:29 -04:00
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2008-01-18 10:54:22 -05:00
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2009-06-09 20:28:34 -04:00
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2009-06-09 16:42:22 -04:00
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2009-06-09 20:28:34 -04:00
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2009-06-10 09:51:32 -04:00
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2009-06-09 16:42:22 -04:00
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2009-06-09 20:28:34 -04:00
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2009-06-09 16:42:22 -04:00
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2008-01-09 09:23:21 -05:00
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2008-06-10 10:40:29 -04:00
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2008-01-09 09:23:21 -05:00
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2008-06-11 21:47:56 -04:00
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2012-04-24 22:59:16 +03:00
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2008-06-11 21:47:56 -04:00
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2008-01-29 16:03:38 -05:00
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2008-06-10 10:40:29 -04:00
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2010-02-28 10:59:11 +00:00
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2008-06-10 10:40:29 -04:00
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2008-01-29 16:03:38 -05:00
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2008-06-11 16:51:38 -04:00
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2008-06-10 10:40:29 -04:00
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2009-04-21 12:38:29 -07:00
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2008-01-29 16:03:38 -05:00
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2008-01-02 10:01:11 -05:00
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2008-06-10 10:40:29 -04:00
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Btrfs: protect fs_info->alloc_start
fs_info->alloc_start is a 64bits variant, can be accessed by
multi-task, but it is not protected strictly, it can be changed
while we are accessing it. On 32bit machine, we will get wrong
value because we access it by two instructions.(In fact, it is
also possible that the same problem happens on the 64bit machine,
because the compiler may split the 64bit operation into two 32bit
operation.)
For example:
Assuming -> alloc_start is 0x0000 0000 0001 0000 at the beginning,
then we remount and set ->alloc_start to 0x0000 0100 0000 0000.
Task0 Task1
load high 32 bits
set high 32 bits
set low 32 bits
load low 32 bits
Task1 will get 0.
This patch fixes this problem by using two locks to protect it
fs_info->chunk_mutex
sb->s_umount
On the read side, we just need get one of these two locks, and on
the write side, we must lock all of them.
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Josef Bacik <jbacik@fusionio.com>
2013-01-29 10:07:33 +00:00
|
|
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2010-02-28 10:59:11 +00:00
|
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|
Btrfs: protect fs_info->alloc_start
fs_info->alloc_start is a 64bits variant, can be accessed by
multi-task, but it is not protected strictly, it can be changed
while we are accessing it. On 32bit machine, we will get wrong
value because we access it by two instructions.(In fact, it is
also possible that the same problem happens on the 64bit machine,
because the compiler may split the 64bit operation into two 32bit
operation.)
For example:
Assuming -> alloc_start is 0x0000 0000 0001 0000 at the beginning,
then we remount and set ->alloc_start to 0x0000 0100 0000 0000.
Task0 Task1
load high 32 bits
set high 32 bits
set low 32 bits
load low 32 bits
Task1 will get 0.
This patch fixes this problem by using two locks to protect it
fs_info->chunk_mutex
sb->s_umount
On the read side, we just need get one of these two locks, and on
the write side, we must lock all of them.
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Josef Bacik <jbacik@fusionio.com>
2013-01-29 10:07:33 +00:00
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2008-06-10 10:40:29 -04:00
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2009-04-21 12:38:29 -07:00
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2008-01-02 10:01:11 -05:00
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2008-07-24 12:16:36 -04:00
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2009-04-02 16:49:40 -04:00
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2009-04-02 16:59:01 -04:00
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2009-04-21 17:40:57 -04:00
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2009-10-14 09:24:59 -04:00
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2010-06-21 14:48:16 -04:00
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2010-11-19 13:40:41 +00:00
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2011-10-03 14:07:49 -04:00
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2011-06-03 09:36:29 -04:00
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2010-09-21 14:21:34 -04:00
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2010-06-21 14:48:16 -04:00
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2010-10-29 15:46:43 -04:00
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2011-02-16 13:10:41 -05:00
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2011-05-24 15:35:30 -04:00
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2012-05-07 06:35:38 -06:00
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2011-05-24 15:35:30 -04:00
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2011-11-03 15:17:42 -04:00
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2012-05-07 06:35:38 -06:00
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2011-11-03 15:17:42 -04:00
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2012-01-16 22:04:48 +02:00
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2011-11-09 13:44:05 +01:00
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2011-10-03 23:22:31 -04:00
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2009-11-07 06:19:16 +00:00
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2007-08-29 09:11:44 -04:00
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2007-12-17 20:14:01 -05:00
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2007-08-29 09:11:44 -04:00
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2009-11-07 06:19:16 +00:00
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2011-10-03 14:07:49 -04:00
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2010-02-25 20:38:35 +00:00
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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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2008-12-02 06:36:09 -05:00
|
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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
|
|
|
|
2011-04-06 07:33:51 +00:00
|
|
|
|
2008-06-10 10:40:29 -04:00
|
|
|
|
|
|
|
|
|
2011-09-14 14:11:21 +08:00
|
|
|
|
2008-06-10 10:40:29 -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
|
|
|
|
2008-06-10 10:40:29 -04:00
|
|
|
|
|
|
|
|
|
2011-07-25 15:55:42 -04:00
|
|
|
|
2008-06-10 10:40:29 -04:00
|
|
|
|
|
|
|
|
|
|
|
|
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|
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|
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|
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|
|
|
|
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|
|
|
2010-12-27 16:43:13 +08:00
|
|
|
|
2008-06-10 10:40:29 -04:00
|
|
|
|
|
|
|
|
|
|
|
|
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|
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|
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|
|
2011-11-08 16:47:55 +02:00
|
|
|
|
2008-06-10 10:40:29 -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
|
|
|
|
|
|
|
|
|
2009-12-14 19:18:38 +00:00
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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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
|
|
|
|
2011-04-06 07:33:51 +00:00
|
|
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|
2008-06-10 10:40:46 -04:00
|
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|
|
2011-09-14 14:11:21 +08:00
|
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|
2008-06-10 10:40:46 -04:00
|
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|
2011-09-14 14:11:21 +08:00
|
|
|
|
2008-06-10 10:40:46 -04:00
|
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|
2011-07-25 15:55:42 -04:00
|
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|
2008-06-10 10:40:46 -04:00
|
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2008-06-10 10:40:29 -04:00
|
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2011-07-25 15:55:42 -04:00
|
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|
2010-12-27 16:43:13 +08:00
|
|
|
|
2008-06-10 10:40:29 -04:00
|
|
|
|
2007-08-29 09:11:44 -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
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2011-11-17 15:40:49 -05:00
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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
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2011-11-17 15:40:49 -05:00
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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
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|
2011-05-14 07:10:51 +00:00
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2010-05-29 09:40:57 +00:00
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2011-05-14 07:10:51 +00:00
|
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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
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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
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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
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|
2010-10-29 15:14:23 -04:00
|
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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
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2010-05-29 09:42:19 +00:00
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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
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2011-07-25 15:40:35 -04:00
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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
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2009-01-05 21:25:51 -05:00
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2008-03-24 15:02:07 -04:00
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2009-01-05 21:25:51 -05:00
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|
2007-03-21 11:12:56 -04:00
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|
2009-01-05 21:25:51 -05:00
|
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|
2011-11-17 15:40:49 -05:00
|
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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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|
2007-06-12 06:35:45 -04:00
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2007-03-21 11:12:56 -04:00
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2007-03-22 12:13:20 -04:00
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2010-12-20 10:56:06 -05:00
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2008-07-21 02:01:56 +05:30
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2007-11-16 11:45:54 -05:00
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2007-03-21 11:12:56 -04:00
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2009-10-13 13:50:18 -04:00
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2008-07-24 12:16:36 -04:00
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2009-09-29 13:51:04 -04:00
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2012-04-05 15:03:02 -04:00
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2011-11-17 01:10:02 -05:00
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2007-03-22 12:13:20 -04:00
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2011-11-17 01:10:02 -05:00
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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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|
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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|
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|
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|
|
2011-11-17 01:00:31 -05:00
|
|
|
|
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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2007-06-12 06:35:45 -04:00
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2007-03-21 11:12:56 -04:00
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2012-01-08 22:15:13 -05:00
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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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2008-02-20 16:11:05 -05:00
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2011-05-26 10:01:56 -06:00
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2011-11-17 01:56:28 -05:00
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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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2011-11-17 15:40:49 -05:00
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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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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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2012-09-14 02:58:07 -06:00
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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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2012-09-14 10:34:40 -04:00
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2011-01-20 06:19:37 +00:00
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2012-09-14 10:34:40 -04:00
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2012-08-24 12:53:03 -06:00
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2007-03-23 10:01:08 -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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2011-03-31 00:44:29 +00:00
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2009-04-02 16:46:06 -04:00
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2009-04-21 12:38:29 -07:00
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2009-04-02 16:46:06 -04:00
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2009-04-21 12:38:29 -07:00
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2009-04-02 16:46:06 -04:00
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2011-03-31 00:44:29 +00:00
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2009-06-10 09:51:32 -04:00
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2009-06-09 20:28:34 -04:00
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2009-04-02 16:46:06 -04:00
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2009-04-02 16:49:40 -04:00
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2009-05-14 13:52:21 -04:00
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2009-04-02 16:59:01 -04:00
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2009-05-14 13:52:21 -04:00
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2009-12-14 22:01:12 +00:00
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2009-04-02 16:46:06 -04:00
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2011-03-31 00:44:29 +00:00
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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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2011-03-31 00:44:29 +00:00
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2011-06-28 15:10:37 +00:00
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2012-01-16 22:04:48 +02:00
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2011-10-03 23:22:31 -04:00
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2009-04-02 16:46:06 -04:00
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2008-05-07 11:43:44 -04:00
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2007-03-21 11:12:56 -04:00
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2011-11-17 15:40:49 -05:00
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2007-08-29 09:11:44 -04:00
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2011-11-17 15:40:49 -05:00
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2007-08-29 09:11:44 -04:00
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2010-11-19 14:59:15 -05:00
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2011-11-17 01:29:09 -05:00
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2010-11-19 14:59:15 -05:00
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2011-09-29 13:11:33 +02:00
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2011-07-25 15:55:42 -04:00
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2012-04-25 21:24:17 +02:00
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2011-07-25 15:55:42 -04:00
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2012-04-25 21:24:17 +02:00
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2011-07-25 15:55:42 -04:00
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2012-04-25 21:24:17 +02:00
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2011-07-25 15:55:42 -04:00
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2012-04-25 21:24:17 +02:00
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2011-07-25 15:55:42 -04:00
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2012-04-25 21:24:17 +02:00
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2011-07-25 15:55:42 -04:00
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2012-04-25 21:24:17 +02:00
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2011-07-25 15:55:42 -04:00
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2012-04-25 21:24:17 +02:00
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2011-07-25 15:55:42 -04:00
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2012-04-25 21:24:17 +02:00
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2011-07-25 15:55:42 -04:00
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2012-04-25 21:24:17 +02:00
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2011-07-25 15:55:42 -04:00
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2012-04-25 21:24:17 +02:00
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2011-07-25 15:55:42 -04:00
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2012-04-25 21:24:17 +02:00
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2011-07-25 15:55:42 -04:00
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2011-11-16 21:43:59 -05:00
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2011-07-25 15:55:42 -04:00
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2011-11-16 21:43:59 -05:00
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2011-09-29 13:11:33 +02:00
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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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2008-06-10 10:40:29 -04:00
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2010-07-26 16:21:33 +04:00
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2011-04-19 14:29:38 +02:00
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2007-08-29 09:11:44 -04:00
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2008-03-24 15:02:07 -04:00
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2010-11-19 14:59:15 -05:00
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2008-12-02 06:36:09 -05:00
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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
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2011-04-06 07:33:51 +00:00
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2007-08-29 09:11:44 -04:00
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2008-12-02 06:36:09 -05:00
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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
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2011-04-06 07:33:51 +00:00
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2011-11-08 19:15:05 +02:00
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2010-07-26 16:21:33 +04:00
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2011-11-08 19:15:05 +02:00
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2008-06-10 10:40:29 -04:00
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2011-07-25 15:55:42 -04:00
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2011-04-19 14:29:38 +02:00
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2008-03-24 15:02:07 -04:00
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2011-07-25 15:55:42 -04:00
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2007-08-29 09:11:44 -04:00
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2010-11-19 14:59:15 -05:00
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2011-11-09 14:41:22 +02:00
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2010-11-19 14:59:15 -05:00
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2011-04-13 15:41:04 +02:00
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2011-11-09 14:41:22 +02:00
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2011-04-13 15:41:04 +02:00
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2008-05-13 13:46:40 -04:00
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2012-06-25 12:55:37 +01:00
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2011-07-25 15:55:42 -04:00
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2007-08-29 09:11:44 -04:00
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2008-11-17 21:11:30 -05:00
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2011-04-13 15:41:04 +02:00
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2011-11-17 01:56:28 -05:00
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2007-08-29 09:11:44 -04:00
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2011-11-17 15:40:49 -05:00
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2008-03-24 15:02:07 -04:00
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2007-08-29 09:11:44 -04:00
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2011-11-17 01:56:28 -05:00
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2011-07-25 15:55:42 -04:00
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2007-08-29 09:11:44 -04:00
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2010-07-26 16:21:33 +04:00
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2007-08-29 09:11:44 -04:00
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2008-11-12 14:34:12 -05:00
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2008-03-24 15:02:07 -04:00
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2011-11-09 14:41:22 +02:00
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2011-04-13 15:41:04 +02:00
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2010-07-26 16:21:33 +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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2012-04-24 22:59:16 +03:00
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2012-11-06 11:43:11 +01:00
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2012-04-24 22:59:16 +03:00
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2013-02-20 23:32:52 -07:00
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2008-11-12 14:34:12 -05:00
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2011-11-17 15:40:49 -05:00
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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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2013-02-20 23:32:52 -07:00
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2009-02-12 09:37:35 -05:00
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2012-03-01 17:24:58 +01:00
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2009-02-12 09:37:35 -05:00
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2012-04-24 22:59:16 +03:00
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2008-11-12 14:34:12 -05:00
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2013-02-20 23:32:52 -07:00
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2008-11-12 14:34:12 -05:00
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2012-11-06 13:15:27 +01:00
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2008-11-12 14:34:12 -05:00
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2012-11-06 13:15:27 +01:00
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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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2012-04-16 06:44:37 +03:00
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2012-03-01 17:24:58 +01:00
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2012-04-16 06:44:37 +03:00
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2008-11-17 21:11:30 -05:00
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2012-10-30 17:16:16 +00:00
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2012-04-16 06:44:37 +03:00
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2012-03-01 17:24:58 +01:00
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2012-04-16 06:44:37 +03:00
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2008-11-12 14:34:12 -05:00
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2011-11-17 15:40:49 -05:00
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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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2010-05-16 10:49:58 -04:00
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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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2012-06-22 12:24:13 -06:00
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2012-11-06 13:15:27 +01:00
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2008-11-12 14:34:12 -05:00
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2013-02-20 23:32:52 -07:00
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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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Btrfs: protect fs_info->alloc_start
fs_info->alloc_start is a 64bits variant, can be accessed by
multi-task, but it is not protected strictly, it can be changed
while we are accessing it. On 32bit machine, we will get wrong
value because we access it by two instructions.(In fact, it is
also possible that the same problem happens on the 64bit machine,
because the compiler may split the 64bit operation into two 32bit
operation.)
For example:
Assuming -> alloc_start is 0x0000 0000 0001 0000 at the beginning,
then we remount and set ->alloc_start to 0x0000 0100 0000 0000.
Task0 Task1
load high 32 bits
set high 32 bits
set low 32 bits
load low 32 bits
Task1 will get 0.
This patch fixes this problem by using two locks to protect it
fs_info->chunk_mutex
sb->s_umount
On the read side, we just need get one of these two locks, and on
the write side, we must lock all of them.
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Josef Bacik <jbacik@fusionio.com>
2013-01-29 10:07:33 +00:00
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2012-03-01 17:24:58 +01:00
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Btrfs: protect fs_info->alloc_start
fs_info->alloc_start is a 64bits variant, can be accessed by
multi-task, but it is not protected strictly, it can be changed
while we are accessing it. On 32bit machine, we will get wrong
value because we access it by two instructions.(In fact, it is
also possible that the same problem happens on the 64bit machine,
because the compiler may split the 64bit operation into two 32bit
operation.)
For example:
Assuming -> alloc_start is 0x0000 0000 0001 0000 at the beginning,
then we remount and set ->alloc_start to 0x0000 0100 0000 0000.
Task0 Task1
load high 32 bits
set high 32 bits
set low 32 bits
load low 32 bits
Task1 will get 0.
This patch fixes this problem by using two locks to protect it
fs_info->chunk_mutex
sb->s_umount
On the read side, we just need get one of these two locks, and on
the write side, we must lock all of them.
Signed-off-by: Miao Xie <miaox@cn.fujitsu.com>
Signed-off-by: Josef Bacik <jbacik@fusionio.com>
2013-01-29 10:07:33 +00:00
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2012-04-24 22:59:16 +03:00
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2012-03-01 17:24:58 +01:00
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2013-02-20 23:32:52 -07:00
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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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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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2011-11-28 16:43:00 +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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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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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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2011-11-28 16:43:00 +08:00
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2012-11-05 18:29:28 +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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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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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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|
2007-10-15 16:15:53 -04:00
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|
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2011-11-17 15:40:49 -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
|
|
|
|
2007-04-19 21:01:03 -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
|
|
|
|
2011-11-17 15:40:49 -05:00
|
|
|
|
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
|
|
|
|
2010-10-14 14:52:27 -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
|
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2010-05-16 10:46:24 -04:00
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|
|
|
2010-10-14 14:52:27 -04:00
|
|
|
|
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
|
|
|
|
|
|
|
|
|
2007-04-19 21:01:03 -04:00
|
|
|
|
2007-10-15 16:15:53 -04:00
|
|
|
|
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
|
|
|
|
2007-04-19 21:01:03 -04:00
|
|
|
|
|
|
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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
|
|
|
|
2011-11-17 15:40:49 -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
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2011-11-17 15:40:49 -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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2011-11-17 15:40:49 -05:00
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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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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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2007-04-19 21:01:03 -04:00
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2007-04-24 11:52:22 -04:00
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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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2011-11-17 01:46:50 -05:00
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2011-11-17 01:22:46 -05:00
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2007-03-21 11:12:56 -04:00
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2010-07-26 16:21:33 +04:00
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2011-11-17 01:22:46 -05:00
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2007-03-21 11:12:56 -04:00
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2008-03-24 15:02:04 -04:00
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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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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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2009-01-16 11:59:08 -05:00
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2008-03-24 15:02:07 -04:00
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2008-12-02 06:36:09 -05:00
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2008-03-24 15:02:07 -04:00
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2012-06-21 16:03:58 -04:00
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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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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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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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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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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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2008-01-22 12:46:56 -05:00
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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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2009-01-10 06:09:52 -08:00
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2008-01-22 12:46:56 -05:00
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2007-03-21 11:12:56 -04:00
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2012-06-05 14:13:12 -04:00
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2012-08-02 10:22:20 -04:00
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2012-06-05 14:13:12 -04:00
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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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2007-03-26 12:00:39 -04:00
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2007-04-02 10:50:19 -04:00
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2007-04-19 21:01:03 -04:00
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2008-11-12 14:34:12 -05:00
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2009-01-10 06:09:52 -08:00
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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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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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2008-03-24 15:02:04 -04:00
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2008-12-02 09:54:17 -05:00
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2008-03-24 15:02:04 -04:00
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2012-05-07 06:35:38 -06:00
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2008-03-24 15:02:04 -04:00
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2007-03-21 11:12:56 -04:00
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2007-04-02 10:50:19 -04:00
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2007-08-29 15:47:34 -04:00
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2012-03-01 14:56:26 +01:00
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2008-01-24 16:13:08 -05:00
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2010-12-17 14:21:50 +08:00
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2008-01-24 16:13:08 -05:00
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2007-11-19 10:22:33 -05:00
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2008-01-24 16:13:08 -05:00
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2012-09-06 04:01:51 -06:00
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2007-11-19 10:22:33 -05:00
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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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2012-09-06 04:01:51 -06:00
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2012-11-26 09:24:43 +00:00
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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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2012-11-21 02:21:28 +00:00
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2012-11-26 09:24:43 +00:00
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2012-11-21 02:21:28 +00:00
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2008-03-24 15:02:04 -04:00
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2008-07-23 12:12:13 -04:00
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2012-03-23 17:14:20 +01:00
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2008-07-23 12:12:13 -04:00
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2007-11-19 10:22:33 -05:00
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2008-03-24 15:02:04 -04:00
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2012-11-21 02:21:28 +00:00
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2012-11-26 09:24:43 +00:00
|
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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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|
|
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|
|
|
2012-09-06 04:01:51 -06:00
|
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|
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|
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|
|
2007-11-19 10:22:33 -05:00
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|
|
2008-01-24 16:13:08 -05:00
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2007-11-19 10:22:33 -05:00
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|
2010-12-17 14:21:50 +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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2007-06-12 06:35:45 -04:00
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|
2012-11-21 02:21:28 +00:00
|
|
|
|
2012-11-26 09:24:43 +00:00
|
|
|
|
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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