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
