Messages in this thread | | | Date | Wed, 18 Nov 2015 00:55:10 +0100 | From | Peter Zijlstra <> | Subject | Re: [PATCH v4 3/3] sched: optimize migration by forcing rmb() and updating to be called once |
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On Wed, Nov 18, 2015 at 08:37:00AM +0900, Byungchul Park wrote:
> Which one do you think to be fixed? The one above migrate_task_rq_fair()? > I wonder if it would be ok even it does not hold pi_lock in > migrate_task_rq_fair(). If you say *no problem*, I will try to fix the > comment.
The one above migrate_task_rq_fair() is obviously broken, as demonstrated by the move_queued_task() case.
Also, pretty much all runnable task migration code will not take pi_lock, see also {pull,push}_{rt,dl}_task().
Note that this is done very much by design, task_rq_lock() is the thing that fully serializes a task's scheduler state. Runnable tasks use rq->lock, waking tasks use pi_lock.
> > I meant, if you call __set_task_cpu() before > > sched_class::migrate_task_rq(), in that case task_rq_lock() will no > > longer fully serialize against set_task_cpu(). > > > > Because once you've called __set_task_cpu(), task_rq_lock() will acquire > > the _other_ rq->lock. And we cannot rely on our rq->lock to serialize > > things. > > I agree with you if migtrate_task_rq() can be serialized by rq->lock > without holding pi_lock. (even though I am still wondering..)
move_queued_task() illustrates this.
> But I thought it was no problem if migrate_task_rq() was serialized only > by pi_lock as the comment above the migrate_task_rq() describes, because > breaking rq->lock does not affect the sericalization by pi_lock.
Right, but per the above, we cannot assume pi_lock is in fact held over this.
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