On Tue, Feb 14, 2023 at 11:48:07AM +0100, Peter Zijlstra wrote: > On Mon, Feb 13, 2023 at 05:51:11PM -0800, Boqun Feng wrote: > > On Mon, Feb 13, 2023 at 05:29:49PM +0100, Peter Zijlstra wrote: > > > On Mon, Feb 13, 2023 at 10:25:59AM -0500, Alan Stern wrote: > > > > On Mon, Feb 13, 2023 at 10:24:13AM +0100, Peter Zijlstra wrote: > > > > > On Sun, Feb 12, 2023 at 10:23:44AM -0500, Alan Stern wrote: > > > > > > Provided it acquires the parent device's lock first, this is > > > > > > utterly safe no matter what order the children are locked in. Try > > > > > > telling that to lockdep! > > > > > > > > > > mutex_lock_next_lock(child->lock, parent->lock) is there to express this > > > > > exact pattern, it allows taking multiple child->lock class locks (in any > > > > > order) provided parent->lock is held. > > > > > > > > Ah, this is news to me. Is this sort of thing documented somewhere? > > > > 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. > > > > No, this isn't quite right, You need at least 3 locks and 2 classes. > > P, C1, C2 > > Then: > > mutex_lock(P) > mutex_lock_next_lock(C1, P) > mutex_lock_next_lock(C2, P) > > And it will accept any order of Cn -- since it assumes that any > multi-lock of Cn will always hold P, therefore the ordering fully given > by P. Ah, right, I was missing the fact that it works with 2 classes... But I think with only one class, the nest_lock() still works, right? In other words, if P and Cn are the same lock class in your example. Also seems I gave a wrong answer to Alan, just to clarify, the following is not a deadlock to lockdep: T1: mutex_lock(P) mutex_lock_next_lock(C1, P) mutex_lock_next_lock(C2, P) mutex_lock(B) T2: mutex_lock(P) mutex_lock(B) mutex_lock_next_lock(C1, P) mutex_lock_next_lock(C2, P) Because of any pair of mutex_lock(L); ... // other locks maybe mutex_lock_nest_lock(M, L); lockdep will not add M into the dependency graph, since it's nested and should be serialized by L. Regards, Boqun
