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().