On 16.12.2016 18:15, Peter Zijlstra wrote: > On Fri, Dec 16, 2016 at 03:19:43PM +0100, Nicolai Hähnle wrote: >> The concern about picking up a handoff that we didn't request is real, >> though it cannot happen in the first iteration. Perhaps this __mutex_trylock >> can be moved to the end of the loop? See below... > > >>>> @@ -728,7 +800,7 @@ __mutex_lock_common(struct mutex *lock, long state, >>>> unsigned int subclass, >>>> * or we must see its unlock and acquire. >>>> */ >>>> if ((first && mutex_optimistic_spin(lock, ww_ctx, use_ww_ctx, >>>> true)) || >>>> - __mutex_trylock(lock, first)) >>>> + __mutex_trylock(lock, use_ww_ctx || first)) >>>> break; >>>> >>>> spin_lock_mutex(&lock->wait_lock, flags); >> >> Change this code to: >> >> acquired = first && >> mutex_optimistic_spin(lock, ww_ctx, use_ww_ctx, >> &waiter); >> spin_lock_mutex(&lock->wait_lock, flags); >> >> if (acquired || >> __mutex_trylock(lock, use_ww_ctx || first)) >> break; > > goto acquired; > > will work lots better.
Wasn't explicit enough, sorry. The idea was to get rid of the acquired label and change things so that all paths exit the loop with wait_lock held. That seems cleaner to me. >> } >> >> This changes the trylock to always be under the wait_lock, but we previously >> had that at the beginning of the loop anyway. > >> It also removes back-to-back >> calls to __mutex_trylock when going through the loop; > > Yeah, I had that explicitly. It allows taking the mutex when > mutex_unlock() is still holding the wait_lock. mutex_optimistic_spin() already calls __mutex_trylock, and for the no-spin case, __mutex_unlock_slowpath() only calls wake_up_q() after releasing the wait_lock. So I don't see the purpose of the back-to-back __mutex_trylocks, especially considering that if the first one succeeds, we immediately take the wait_lock anyway. Nicolai >> and for the first >> iteration, there is a __mutex_trylock under wait_lock already before adding >> ourselves to the wait list. > > Correct. >
