Re: [PATCH]sched: stop hrtick timer if running task is switchingfrom fair scheduling class to another
From: Oleg Nesterov
Date: Mon Dec 26 2011 - 10:48:51 EST
On 12/23, Kirill Tkhai wrote:
> [PATCH]sched: stop hrtick timer if running task is switching from fair
> scheduling class to another
> We have to stop hrtick timer to avoid excess interrupt. Not-fair tasks
> are not interested in fair's hrtick. RT class uses its own fixed
> timeslice (in case of RR), which doesn't depend on current value of hrtick
Well. I shouldn't try to comment this patch, I do not really understand
But since nobody else replies...
> @@ -5271,6 +5271,13 @@ static void switched_from_fair(struct rq *rq,
> struct task_struct *p)
> place_entity(cfs_rq, se, 0);
> se->vruntime -= cfs_rq->min_vruntime;
> + /*
> + * Other scheduling classes are not interested in fair's hrtick timer.
> + */
> + if (task_current(rq, p) && sched_feat(HRTICK))
> + hrtick_clear(rq);
May be... but in this case, perhaps instead we should teach
dequeue_task_fair() or put_prev_task_fair() to do this. Then we can
probably remove __schedule()->hrtick_clear().
I simply can't understand the current hrtick logic. For example,
why dequeue_task_fair() does hrtick_update() ? OK, probably
because "nr_running < sched_nr_latency" can become true. But at
least this doesn't make sense to me when p == rq->curr, say,
Hmm. In any case, how it is possible to do hrtick_start() with
rq->lock held? hrtimer_restart() may want to wakeup_softirqd().
Peter, Ingo, does this code really work? SCHED_FEAT(HRTICK) == 0
by default, and afaics you can't change it without CONFIG_SCHED_DEBUG.
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