On Tue, Feb 14, 2023 at 01:27:33PM +0800, Hillf Danton wrote: > On Mon, 13 Feb 2023 18:09:16 -0800 Boqun Feng <boqun.feng@xxxxxxxxx> > > On Mon, Feb 13, 2023 at 09:03:14PM -0500, Alan Stern wrote: > > > On Mon, Feb 13, 2023 at 05:51:11PM -0800, Boqun Feng wrote: > > > > Basically if you have two lock instances A and B with the same class, > > > > and you know that locking ordering is always A -> B, then you can do > > > > > > > > mutex_lock(A); > > > > mutex_lock_nest_lock(B, A); // lock B. > > > > > > > > to tell the lockdep this is not deadlock, plus lockdep will treat the > > > > acquisition of A and the precondition of acquisition B, so the following > > > > is not a deadlock as well: > > > > > > > > T1: > > > > mutex_lock(A); > > > > mutex_lock(C); > > > > mutex_lock_nest_lock(B, A); > > > > > > > > T2: > > > > mutex_lock(A); > > > > mutex_lock_nest_lock(B, A); > > > > mutex_lock(C); > > > > > > Why isn't this treated as a deadlock? It sure looks like a deadlock to > > > me. Is this an example where lockdep just doesn't get the right answer? > > Syzbot reported deadlock[1] with A ignored. > > [1] https://lore.kernel.org/linux-mm/20230130073136.59-1-hdanton@xxxxxxxx/ > Right, I think that's a false positive, however it's not related to mutex_lock_nest_lock(). Anyway mutex_lock_nest_lock() cannot help that case since these are three different lock class. Actually, reading the code again, I think I made a mistake, for mutex_lock_nest_lock(), the following *is* a deadlock to lockdep: T1: mutex_lock(A); mutex_lock(C); mutex_lock_nest_lock(B, A); T2: mutex_lock(A); mutex_lock_nest_lock(B, A); mutex_lock(C); and this *is not* a deadlock to lockdep: T1: mutex_lock(A); mutex_lock_nest_lock(C, A); mutex_lock_nest_lock(B, A); T2: mutex_lock(A); mutex_lock_nest_lock(B, A); mutex_lock_nest_lock(C, A); The current semantics of _nest_lock() is tricky, it only provides the "nest" effect if it is the next lock acquired after the "parent" lock. Maybe we can change and make it clear a little bit to make it more useful. Ah, actually someone found it 7 years ago: https://lore.kernel.org/lkml/20150810095247.GA4606@xxxxxxxxxxxxxxxxxxxxxxx/ ;-) Regards, Boqun > > Because A serializes B and C, so that particular piece of code doesn't > > cause deadlock. Note that you can still use you normal mutex_lock() for > > B, so if there is more code: > > > > T3: > > mutex_lock(C); > > mutex_lock(B); > > > > lockdep will report deadlock. > > > > Regards, > > Boqun > > > > > Alan Stern
