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

Morten Rasmussen
2016-06-07T14:24:28Z

On Wed, Jun 01, 2016 at 09:59:20PM +0200, Peter Zijlstra wrote:
> On Mon, May 23, 2016 at 11:58:46AM +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@redhat.com>
> > cc: Peter Zijlstra <peterz@infradead.org>
> >
> > Signed-off-by: Morten Rasmussen <morten.rasmussen@arm.com>
> > ---
> > 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));
> > + }
>
> superfluous brackets, also isn't @this_cpu supposed to be part of the
> sched_group_cpu(group) mask?

I'll kill the brackets.

IIUC, @this_cpu is never part of sched_group_cpus(group).
find_idlest_group() never returns the local group, so if we manage to
get to find_idlest_cpu(), the group would not contain @this_cpu.