Re: [PATCH] sched/deadline: Remove unnecessary condition in push_dl_task()

[Steven Rostedt]

On Wed, 15 Feb 2017 10:47:49 +0000
Juri Lelli <juri.lelli@xxxxxxx> wrote:

> [+Steve, Luca]
> 
> Hi,
> 
> On 15/02/17 14:11, Byungchul Park wrote:
> > Once pick_next_pushable_dl_task(rq) return a task, it guarantees that
> > the task's cpu is rq->cpu, so task_cpu(next_task) is always rq->cpu if
> > task == next_task. Remove a redundant condition and make code simpler.
> > 
> > Signed-off-by: Byungchul Park <byungchul.park@xxxxxxx>
> > ---
> >  kernel/sched/deadline.c | 2 +-
> >  1 file changed, 1 insertion(+), 1 deletion(-)
> > 
> > diff --git a/kernel/sched/deadline.c b/kernel/sched/deadline.c
> > index 27737f3..ad8d577 100644
> > --- a/kernel/sched/deadline.c
> > +++ b/kernel/sched/deadline.c
> > @@ -1483,7 +1483,7 @@ static int push_dl_task(struct rq *rq)
> >  		 * then possible that next_task has migrated.
> >  		 */
> >  		task = pick_next_pushable_dl_task(rq);
> > -		if (task_cpu(next_task) == rq->cpu && task == next_task) {
> > +		if (task == next_task) {  
> 
> Seems a sensible optimization to me. Actually, we are doing the same for
> rt.c; Steve, Peter, do you think we should optimize that as well?
> 

Are we doing the same for push_rt_task()? I don't see it, and I don't
see it in tip/sched/core either. What I have is:

  if (task_cpu(next_task) == rq->cpu && task == next_task) {

But that said, I believe this patch is correct, and we should change
rt.c as well.


  task = pick_next_pushable_dl_task(rq);

Which has:

  BUG_ON(rq->cpu != task_cpu(task))

when it returns a task other than NULL. Which means that task_cpu(task)
must be rq->cpu. Then if task == next_task, then task_cpu(next_task)
must be rq->cpu as well.

Reviewed-by: Steven Rostedt (VMware) <rostedt@xxxxxxxxxxx>

Mind fixing rt.c if it hasn't been fixed already.

-- Steve