Re: [PATCH v2 05/11] locking/ww_mutex: Add waiters in stamp order

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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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