Re: [PATCH 04/16] sched/fair: Optimize find_idlest_cpu() when there is no choice

[Morten Rasmussen]

On Tue, May 24, 2016 at 08:29:05AM +0200, Mike Galbraith wrote:
> On Mon, 2016-05-23 at 11:58 +0100, Morten Rasmussen wrote:
> > In the current find_idlest_group()/find_idlest_cpu() search we end up
> > calling find_idlest_cpu() in a sched_group containing only one cpu in
> > the end. Checking idle-states becomes pointless when there is no
> > alternative, so bail out instead.
> > 
> > cc: Ingo Molnar <mingo@xxxxxxxxxx>
> > cc: Peter Zijlstra <peterz@xxxxxxxxxxxxx>
> > 
> > Signed-off-by: Morten Rasmussen <morten.rasmussen@xxxxxxx>
> > ---
> >  kernel/sched/fair.c | 5 +++++
> >  1 file changed, 5 insertions(+)
> > 
> > diff --git a/kernel/sched/fair.c b/kernel/sched/fair.c
> > index 0fe3020..564215d 100644
> > --- a/kernel/sched/fair.c
> > +++ b/kernel/sched/fair.c
> > @@ -5155,6 +5155,11 @@ find_idlest_cpu(struct sched_group *group, struct task_struct *p, int this_cpu)
> >  > 	> int shallowest_idle_cpu = -1;
> >  > 	> int i;
> >  
> > +> 	> /* Check if we have any choice */
> > +> 	> if (group->group_weight == 1) {
> > +> 	> 	> return cpumask_first(sched_group_cpus(group));
> > +> 	> }
> > +
> 
> Hm, if task isn't allowed there, too bad?

Is that possible for single-cpu groups? I thought we skipped groups with
no cpus allowed in find_idlest_group():

                /* Skip over this group if it has no CPUs allowed */
                if (!cpumask_intersects(sched_group_cpus(group),
                                        tsk_cpus_allowed(p)))
                        continue;

Since the group has at least one cpu allowed and only contains one cpu,
that cpu must be allowed. No?