On Wed, Aug 08, 2018 at 04:09:12PM -0400, J. Bruce Fields wrote: > On Wed, Aug 08, 2018 at 03:54:45PM -0400, J. Bruce Fields wrote: > > On Wed, Aug 08, 2018 at 11:51:07AM +1000, NeilBrown wrote: > > > If you have a many-core machine, and have many threads all wanting to > > > briefly lock a give file (udev is known to do this), you can get quite > > > poor performance. > > > > > > When one thread releases a lock, it wakes up all other threads that > > > are waiting (classic thundering-herd) - one will get the lock and the > > > others go to sleep. > > > When you have few cores, this is not very noticeable: by the time the > > > 4th or 5th thread gets enough CPU time to try to claim the lock, the > > > earlier threads have claimed it, done what was needed, and released. > > > With 50+ cores, the contention can easily be measured. > > > > > > This patchset creates a tree of pending lock request in which siblings > > > don't conflict and each lock request does conflict with its parent. > > > When a lock is released, only requests which don't conflict with each > > > other a woken. > > > > Are you sure you aren't depending on the (incorrect) assumption that "X > > blocks Y" is a transitive relation? > > > > OK I should be able to answer that question myself, my patience for > > code-reading is at a real low this afternoon.... > > In other words, is there the possibility of a tree of, say, exclusive > locks with (offset, length) like: > > (0, 2) waiting on (1, 2) waiting on (2, 2) waiting on (0, 4) > > and when waking (0, 4) you could wake up (2, 2) but not (0, 2), leaving > a process waiting without there being an actual conflict. After batting it back and forth with Jeff on IRC.... So do I understand right that when we wake a waiter, we leave its own tree of waiters intact, and when it wakes if it finds a conflict it just adds it lock (with tree of waiters) in to the tree of the conflicting lock? If so then yes I think that depends on the transitivity assumption--you're assuming that finding a conflict between the root of the tree and a lock proves that all the other members of the tree also conflict. So maybe this example works. (All locks are exclusive and written (offset, length), X->Y means X is waiting on Y.) process acquires (0,3) 2nd process requests (1,2), is put to sleep. 3rd process requests (0,2), is put to sleep. The tree of waiters now looks like (0,2)->(1,2)->(0,3) (0,3) is unlocked. A 4th process races in and locks (2,2). The 2nd process wakes up, sees this new conflict, and waits on (2,2). Now the tree looks like (0,2)->(1,2)->(2,2), and (0,2) is waiting for no reason. ? --b.
