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

[Nicolai Hähnle]

Hi Peter and Chris,

(trying to combine the handoff discussion here)

On 06.12.2016 17:55, Peter Zijlstra wrote:
> On Thu, Dec 01, 2016 at 03:06:48PM +0100, Nicolai Hähnle wrote:
>> @@ -693,8 +748,12 @@ __mutex_lock_common(struct mutex *lock, long state, unsigned int subclass,
>>  		 * mutex_unlock() handing the lock off to us, do a trylock
>>  		 * before testing the error conditions to make sure we pick up
>>  		 * the handoff.
>> +		 *
>> +		 * For w/w locks, we always need to do this even if we're not
>> +		 * currently the first waiter, because we may have been the
>> +		 * first waiter during the unlock.
>>  		 */
>> -		if (__mutex_trylock(lock, first))
>> +		if (__mutex_trylock(lock, use_ww_ctx || first))
>>  			goto acquired;
>
> So I'm somewhat uncomfortable with this. The point is that with the
> .handoff logic it is very easy to accidentally allow:
>
> 	mutex_lock(&a);
> 	mutex_lock(&a);
>
> And I'm not sure this doesn't make that happen for ww_mutexes. We get to
> this __mutex_trylock() without first having blocked.

Okay, took me a while, but I see the problem. If we have:

	ww_mutex_lock(&a, NULL);
	ww_mutex_lock(&a, ctx);

then it's possible that another currently waiting task sets the HANDOFF 
flag between those calls and we'll allow the second ww_mutex_lock to go 
through.

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...


>
>
>>  		/*
>> @@ -716,7 +775,20 @@ __mutex_lock_common(struct mutex *lock, long state, unsigned int subclass,
>>  		spin_unlock_mutex(&lock->wait_lock, flags);
>>  		schedule_preempt_disabled();
>>
>> -		if (!first && __mutex_waiter_is_first(lock, &waiter)) {
>> +		if (use_ww_ctx && ww_ctx) {
>> +			/*
>> +			 * Always re-check whether we're in first position. We
>> +			 * don't want to spin if another task with a lower
>> +			 * stamp has taken our position.
>> +			 *
>> +			 * We also may have to set the handoff flag again, if
>> +			 * our position at the head was temporarily taken away.
>> +			 */
>> +			first = __mutex_waiter_is_first(lock, &waiter);
>> +
>> +			if (first)
>> +				__mutex_set_flag(lock, MUTEX_FLAG_HANDOFF);
>> +		} else if (!first && __mutex_waiter_is_first(lock, &waiter)) {
>>  			first = true;
>>  			__mutex_set_flag(lock, MUTEX_FLAG_HANDOFF);
>>  		}
>
> So the point is that !ww_ctx entries are 'skipped' during the insertion
> and therefore, if one becomes first, it must stay first?

Yes. Actually, it should be possible to replace all the cases of 
use_ww_ctx || first with ww_ctx. Similarly, all cases of use_ww_ctx && 
ww_ctx could be replaced by just ww_ctx.

>
>> @@ -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;
	}

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; and 
for the first iteration, there is a __mutex_trylock under wait_lock 
already before adding ourselves to the wait list.

What do you think?

Nicolai