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.
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