Re: [PATCH 4/4] sched: Idle task shortcut optimization

[Peter Zijlstra <peterz@infradead.org>]

On Fri, Jan 17, 2014 at 04:06:45PM +0100, Daniel Lezcano wrote:
> >I suppose we could write something like:
> >
> >struct task_struct *pick_next_task(struct rq *rq, struct task_struct *prev)
> >{
> >	const struct sched_class *class;
> >	struct task_struct *p;
> >
> >again:
> >	if (likely(rq->nr_running)) {
> >
> >		if (likely(rq->nr_running == rq->cfs.h_nr_running))
> >			return fair_sched_class.pick_next_task(rq, prev);
> >
> >		for_each_class(class) {
> >			p = class->pick_next_task(rq, prev);
> >			if (p)
> >				return p;
> >		}
> >	}
> >
> >	if (idle_balance(rq))
> >		goto again;
> >
> >	rq->idle_stamp = rq_clock(rq);
> >
> >	return idle_sched_class.pick_next_task(rq, prev);
> >}
> >
> >Which keeps idle_balance() before put_prev_task(), and by using
> >idle_sched_clas.pick_next_task() doesn't rape the idle class interface
> >like you did :-)
> 
> But put_prev_task is called before pick_next_task, so idle_balance() is
> called after now, no  ?

No, put_prev_task() is called by the pick_next_task() that returns a
task. So in the idle case above, the idle_sched_class.pick_next_task()
will do the required put_prev_task().