Messages in this thread | | | Date | Fri, 17 Jan 2014 16:23:10 +0100 | From | Peter Zijlstra <> | Subject | Re: [PATCH 4/4] sched: Idle task shortcut optimization |
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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().
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