linux/kernel/rtmutex.c

1046 lines
26 KiB

rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
[PATCH] clean up and remove some extra spinlocks from rtmutex Oleg brought up some interesting points about grabbing the pi_lock for some protections. In this discussion, I realized that there are some places that the pi_lock is being grabbed when it really wasn't necessary. Also this patch does a little bit of clean up. This patch basically does three things: 1) renames the "boost" variable to "chain_walk". Since it is used in the debugging case when it isn't going to be boosted. It better describes what the test is going to do if it succeeds. 2) moves get_task_struct to just before the unlocking of the wait_lock. This removes duplicate code, and makes it a little easier to read. The owner wont go away while either the pi_lock or the wait_lock are held. 3) removes the pi_locking and owner blocked checking completely from the debugging case. This is because the grabbing the lock and doing the check, then releasing the lock is just so full of races. It's just as good to go ahead and call the pi_chain_walk function, since after releasing the lock the owner can then block anyway, and we would have missed that. For the debug case, we really do want to do the chain walk to test for deadlocks anyway. [oleg@tv-sign.ru: more of the same] Signed-of-by: Steven Rostedt <rostedt@goodmis.org> Cc: Ingo Molnar <mingo@elte.hu> Cc: Thomas Gleixner <tglx@linutronix.de> Cc: Oleg Nesterov <oleg@tv-sign.ru> Cc: Esben Nielsen <nielsen.esben@googlemail.com> Signed-off-by: Andrew Morton <akpm@osdl.org> Signed-off-by: Linus Torvalds <torvalds@osdl.org>
2006-09-29 01:59:44 -07:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
[PATCH] clean up and remove some extra spinlocks from rtmutex Oleg brought up some interesting points about grabbing the pi_lock for some protections. In this discussion, I realized that there are some places that the pi_lock is being grabbed when it really wasn't necessary. Also this patch does a little bit of clean up. This patch basically does three things: 1) renames the "boost" variable to "chain_walk". Since it is used in the debugging case when it isn't going to be boosted. It better describes what the test is going to do if it succeeds. 2) moves get_task_struct to just before the unlocking of the wait_lock. This removes duplicate code, and makes it a little easier to read. The owner wont go away while either the pi_lock or the wait_lock are held. 3) removes the pi_locking and owner blocked checking completely from the debugging case. This is because the grabbing the lock and doing the check, then releasing the lock is just so full of races. It's just as good to go ahead and call the pi_chain_walk function, since after releasing the lock the owner can then block anyway, and we would have missed that. For the debug case, we really do want to do the chain walk to test for deadlocks anyway. [oleg@tv-sign.ru: more of the same] Signed-of-by: Steven Rostedt <rostedt@goodmis.org> Cc: Ingo Molnar <mingo@elte.hu> Cc: Thomas Gleixner <tglx@linutronix.de> Cc: Oleg Nesterov <oleg@tv-sign.ru> Cc: Esben Nielsen <nielsen.esben@googlemail.com> Signed-off-by: Andrew Morton <akpm@osdl.org> Signed-off-by: Linus Torvalds <torvalds@osdl.org>
2006-09-29 01:59:44 -07:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
[PATCH] clean up and remove some extra spinlocks from rtmutex Oleg brought up some interesting points about grabbing the pi_lock for some protections. In this discussion, I realized that there are some places that the pi_lock is being grabbed when it really wasn't necessary. Also this patch does a little bit of clean up. This patch basically does three things: 1) renames the "boost" variable to "chain_walk". Since it is used in the debugging case when it isn't going to be boosted. It better describes what the test is going to do if it succeeds. 2) moves get_task_struct to just before the unlocking of the wait_lock. This removes duplicate code, and makes it a little easier to read. The owner wont go away while either the pi_lock or the wait_lock are held. 3) removes the pi_locking and owner blocked checking completely from the debugging case. This is because the grabbing the lock and doing the check, then releasing the lock is just so full of races. It's just as good to go ahead and call the pi_chain_walk function, since after releasing the lock the owner can then block anyway, and we would have missed that. For the debug case, we really do want to do the chain walk to test for deadlocks anyway. [oleg@tv-sign.ru: more of the same] Signed-of-by: Steven Rostedt <rostedt@goodmis.org> Cc: Ingo Molnar <mingo@elte.hu> Cc: Thomas Gleixner <tglx@linutronix.de> Cc: Oleg Nesterov <oleg@tv-sign.ru> Cc: Esben Nielsen <nielsen.esben@googlemail.com> Signed-off-by: Andrew Morton <akpm@osdl.org> Signed-off-by: Linus Torvalds <torvalds@osdl.org>
2006-09-29 01:59:44 -07:00
[PATCH] clean up and remove some extra spinlocks from rtmutex Oleg brought up some interesting points about grabbing the pi_lock for some protections. In this discussion, I realized that there are some places that the pi_lock is being grabbed when it really wasn't necessary. Also this patch does a little bit of clean up. This patch basically does three things: 1) renames the "boost" variable to "chain_walk". Since it is used in the debugging case when it isn't going to be boosted. It better describes what the test is going to do if it succeeds. 2) moves get_task_struct to just before the unlocking of the wait_lock. This removes duplicate code, and makes it a little easier to read. The owner wont go away while either the pi_lock or the wait_lock are held. 3) removes the pi_locking and owner blocked checking completely from the debugging case. This is because the grabbing the lock and doing the check, then releasing the lock is just so full of races. It's just as good to go ahead and call the pi_chain_walk function, since after releasing the lock the owner can then block anyway, and we would have missed that. For the debug case, we really do want to do the chain walk to test for deadlocks anyway. [oleg@tv-sign.ru: more of the same] Signed-of-by: Steven Rostedt <rostedt@goodmis.org> Cc: Ingo Molnar <mingo@elte.hu> Cc: Thomas Gleixner <tglx@linutronix.de> Cc: Oleg Nesterov <oleg@tv-sign.ru> Cc: Esben Nielsen <nielsen.esben@googlemail.com> Signed-off-by: Andrew Morton <akpm@osdl.org> Signed-off-by: Linus Torvalds <torvalds@osdl.org>
2006-09-29 01:59:44 -07:00
[PATCH] clean up and remove some extra spinlocks from rtmutex Oleg brought up some interesting points about grabbing the pi_lock for some protections. In this discussion, I realized that there are some places that the pi_lock is being grabbed when it really wasn't necessary. Also this patch does a little bit of clean up. This patch basically does three things: 1) renames the "boost" variable to "chain_walk". Since it is used in the debugging case when it isn't going to be boosted. It better describes what the test is going to do if it succeeds. 2) moves get_task_struct to just before the unlocking of the wait_lock. This removes duplicate code, and makes it a little easier to read. The owner wont go away while either the pi_lock or the wait_lock are held. 3) removes the pi_locking and owner blocked checking completely from the debugging case. This is because the grabbing the lock and doing the check, then releasing the lock is just so full of races. It's just as good to go ahead and call the pi_chain_walk function, since after releasing the lock the owner can then block anyway, and we would have missed that. For the debug case, we really do want to do the chain walk to test for deadlocks anyway. [oleg@tv-sign.ru: more of the same] Signed-of-by: Steven Rostedt <rostedt@goodmis.org> Cc: Ingo Molnar <mingo@elte.hu> Cc: Thomas Gleixner <tglx@linutronix.de> Cc: Oleg Nesterov <oleg@tv-sign.ru> Cc: Esben Nielsen <nielsen.esben@googlemail.com> Signed-off-by: Andrew Morton <akpm@osdl.org> Signed-off-by: Linus Torvalds <torvalds@osdl.org>
2006-09-29 01:59:44 -07:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
[PATCH] clean up and remove some extra spinlocks from rtmutex Oleg brought up some interesting points about grabbing the pi_lock for some protections. In this discussion, I realized that there are some places that the pi_lock is being grabbed when it really wasn't necessary. Also this patch does a little bit of clean up. This patch basically does three things: 1) renames the "boost" variable to "chain_walk". Since it is used in the debugging case when it isn't going to be boosted. It better describes what the test is going to do if it succeeds. 2) moves get_task_struct to just before the unlocking of the wait_lock. This removes duplicate code, and makes it a little easier to read. The owner wont go away while either the pi_lock or the wait_lock are held. 3) removes the pi_locking and owner blocked checking completely from the debugging case. This is because the grabbing the lock and doing the check, then releasing the lock is just so full of races. It's just as good to go ahead and call the pi_chain_walk function, since after releasing the lock the owner can then block anyway, and we would have missed that. For the debug case, we really do want to do the chain walk to test for deadlocks anyway. [oleg@tv-sign.ru: more of the same] Signed-of-by: Steven Rostedt <rostedt@goodmis.org> Cc: Ingo Molnar <mingo@elte.hu> Cc: Thomas Gleixner <tglx@linutronix.de> Cc: Oleg Nesterov <oleg@tv-sign.ru> Cc: Esben Nielsen <nielsen.esben@googlemail.com> Signed-off-by: Andrew Morton <akpm@osdl.org> Signed-off-by: Linus Torvalds <torvalds@osdl.org>
2006-09-29 01:59:44 -07:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
[PATCH] clean up and remove some extra spinlocks from rtmutex Oleg brought up some interesting points about grabbing the pi_lock for some protections. In this discussion, I realized that there are some places that the pi_lock is being grabbed when it really wasn't necessary. Also this patch does a little bit of clean up. This patch basically does three things: 1) renames the "boost" variable to "chain_walk". Since it is used in the debugging case when it isn't going to be boosted. It better describes what the test is going to do if it succeeds. 2) moves get_task_struct to just before the unlocking of the wait_lock. This removes duplicate code, and makes it a little easier to read. The owner wont go away while either the pi_lock or the wait_lock are held. 3) removes the pi_locking and owner blocked checking completely from the debugging case. This is because the grabbing the lock and doing the check, then releasing the lock is just so full of races. It's just as good to go ahead and call the pi_chain_walk function, since after releasing the lock the owner can then block anyway, and we would have missed that. For the debug case, we really do want to do the chain walk to test for deadlocks anyway. [oleg@tv-sign.ru: more of the same] Signed-of-by: Steven Rostedt <rostedt@goodmis.org> Cc: Ingo Molnar <mingo@elte.hu> Cc: Thomas Gleixner <tglx@linutronix.de> Cc: Oleg Nesterov <oleg@tv-sign.ru> Cc: Esben Nielsen <nielsen.esben@googlemail.com> Signed-off-by: Andrew Morton <akpm@osdl.org> Signed-off-by: Linus Torvalds <torvalds@osdl.org>
2006-09-29 01:59:44 -07:00
[PATCH] clean up and remove some extra spinlocks from rtmutex Oleg brought up some interesting points about grabbing the pi_lock for some protections. In this discussion, I realized that there are some places that the pi_lock is being grabbed when it really wasn't necessary. Also this patch does a little bit of clean up. This patch basically does three things: 1) renames the "boost" variable to "chain_walk". Since it is used in the debugging case when it isn't going to be boosted. It better describes what the test is going to do if it succeeds. 2) moves get_task_struct to just before the unlocking of the wait_lock. This removes duplicate code, and makes it a little easier to read. The owner wont go away while either the pi_lock or the wait_lock are held. 3) removes the pi_locking and owner blocked checking completely from the debugging case. This is because the grabbing the lock and doing the check, then releasing the lock is just so full of races. It's just as good to go ahead and call the pi_chain_walk function, since after releasing the lock the owner can then block anyway, and we would have missed that. For the debug case, we really do want to do the chain walk to test for deadlocks anyway. [oleg@tv-sign.ru: more of the same] Signed-of-by: Steven Rostedt <rostedt@goodmis.org> Cc: Ingo Molnar <mingo@elte.hu> Cc: Thomas Gleixner <tglx@linutronix.de> Cc: Oleg Nesterov <oleg@tv-sign.ru> Cc: Esben Nielsen <nielsen.esben@googlemail.com> Signed-off-by: Andrew Morton <akpm@osdl.org> Signed-off-by: Linus Torvalds <torvalds@osdl.org>
2006-09-29 01:59:44 -07:00
[PATCH] clean up and remove some extra spinlocks from rtmutex Oleg brought up some interesting points about grabbing the pi_lock for some protections. In this discussion, I realized that there are some places that the pi_lock is being grabbed when it really wasn't necessary. Also this patch does a little bit of clean up. This patch basically does three things: 1) renames the "boost" variable to "chain_walk". Since it is used in the debugging case when it isn't going to be boosted. It better describes what the test is going to do if it succeeds. 2) moves get_task_struct to just before the unlocking of the wait_lock. This removes duplicate code, and makes it a little easier to read. The owner wont go away while either the pi_lock or the wait_lock are held. 3) removes the pi_locking and owner blocked checking completely from the debugging case. This is because the grabbing the lock and doing the check, then releasing the lock is just so full of races. It's just as good to go ahead and call the pi_chain_walk function, since after releasing the lock the owner can then block anyway, and we would have missed that. For the debug case, we really do want to do the chain walk to test for deadlocks anyway. [oleg@tv-sign.ru: more of the same] Signed-of-by: Steven Rostedt <rostedt@goodmis.org> Cc: Ingo Molnar <mingo@elte.hu> Cc: Thomas Gleixner <tglx@linutronix.de> Cc: Oleg Nesterov <oleg@tv-sign.ru> Cc: Esben Nielsen <nielsen.esben@googlemail.com> Signed-off-by: Andrew Morton <akpm@osdl.org> Signed-off-by: Linus Torvalds <torvalds@osdl.org>
2006-09-29 01:59:44 -07:00
[PATCH] clean up and remove some extra spinlocks from rtmutex Oleg brought up some interesting points about grabbing the pi_lock for some protections. In this discussion, I realized that there are some places that the pi_lock is being grabbed when it really wasn't necessary. Also this patch does a little bit of clean up. This patch basically does three things: 1) renames the "boost" variable to "chain_walk". Since it is used in the debugging case when it isn't going to be boosted. It better describes what the test is going to do if it succeeds. 2) moves get_task_struct to just before the unlocking of the wait_lock. This removes duplicate code, and makes it a little easier to read. The owner wont go away while either the pi_lock or the wait_lock are held. 3) removes the pi_locking and owner blocked checking completely from the debugging case. This is because the grabbing the lock and doing the check, then releasing the lock is just so full of races. It's just as good to go ahead and call the pi_chain_walk function, since after releasing the lock the owner can then block anyway, and we would have missed that. For the debug case, we really do want to do the chain walk to test for deadlocks anyway. [oleg@tv-sign.ru: more of the same] Signed-of-by: Steven Rostedt <rostedt@goodmis.org> Cc: Ingo Molnar <mingo@elte.hu> Cc: Thomas Gleixner <tglx@linutronix.de> Cc: Oleg Nesterov <oleg@tv-sign.ru> Cc: Esben Nielsen <nielsen.esben@googlemail.com> Signed-off-by: Andrew Morton <akpm@osdl.org> Signed-off-by: Linus Torvalds <torvalds@osdl.org>
2006-09-29 01:59:44 -07:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00
rtmutex: Simplify PI algorithm and make highest prio task get lock In current rtmutex, the pending owner may be boosted by the tasks in the rtmutex's waitlist when the pending owner is deboosted or a task in the waitlist is boosted. This boosting is unrelated, because the pending owner does not really take the rtmutex. It is not reasonable. Example. time1: A(high prio) onwers the rtmutex. B(mid prio) and C (low prio) in the waitlist. time2 A release the lock, B becomes the pending owner A(or other high prio task) continues to run. B's prio is lower than A, so B is just queued at the runqueue. time3 A or other high prio task sleeps, but we have passed some time The B and C's prio are changed in the period (time2 ~ time3) due to boosting or deboosting. Now C has the priority higher than B. ***Is it reasonable that C has to boost B and help B to get the rtmutex? NO!! I think, it is unrelated/unneed boosting before B really owns the rtmutex. We should give C a chance to beat B and win the rtmutex. This is the motivation of this patch. This patch *ensures* only the top waiter or higher priority task can take the lock. How? 1) we don't dequeue the top waiter when unlock, if the top waiter is changed, the old top waiter will fail and go to sleep again. 2) when requiring lock, it will get the lock when the lock is not taken and: there is no waiter OR higher priority than waiters OR it is top waiter. 3) In any time, the top waiter is changed, the top waiter will be woken up. The algorithm is much simpler than before, no pending owner, no boosting for pending owner. Other advantage of this patch: 1) The states of a rtmutex are reduced a half, easier to read the code. 2) the codes become shorter. 3) top waiter is not dequeued until it really take the lock: they will retain FIFO when it is stolen. Not advantage nor disadvantage 1) Even we may wakeup multiple waiters(any time when top waiter changed), we hardly cause "thundering herd", the number of wokenup task is likely 1 or very little. 2) two APIs are changed. rt_mutex_owner() will not return pending owner, it will return NULL when the top waiter is going to take the lock. rt_mutex_next_owner() always return the top waiter. will not return NULL if we have waiters because the top waiter is not dequeued. I have fixed the code that use these APIs. need updated after this patch is accepted 1) Document/* 2) the testcase scripts/rt-tester/t4-l2-pi-deboost.tst Signed-off-by: Lai Jiangshan <laijs@cn.fujitsu.com> LKML-Reference: <4D3012D5.4060709@cn.fujitsu.com> Reviewed-by: Steven Rostedt <rostedt@goodmis.org> Signed-off-by: Steven Rostedt <rostedt@goodmis.org>
2011-01-14 17:09:41 +08:00