sched/fair: Optimize find_idlest_cpu() when there is no choice

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.

Signed-off-by: Morten Rasmussen <morten.rasmussen@arm.com>
Signed-off-by: Peter Zijlstra (Intel) <peterz@infradead.org>
Cc: Linus Torvalds <torvalds@linux-foundation.org>
Cc: Mike Galbraith <efault@gmx.de>
Cc: Peter Zijlstra <peterz@infradead.org>
Cc: Thomas Gleixner <tglx@linutronix.de>
Cc: dietmar.eggemann@arm.com
Cc: linux-kernel@vger.kernel.org
Cc: mgalbraith@suse.de
Cc: vincent.guittot@linaro.org
Cc: yuyang.du@intel.com
Link: http://lkml.kernel.org/r/1466615004-3503-4-git-send-email-morten.rasmussen@arm.com
Signed-off-by: Ingo Molnar <mingo@kernel.org>
This commit is contained in:
Morten Rasmussen 2016-06-22 18:03:14 +01:00 committed by Ingo Molnar
parent 772bd008cd
commit eaecf41f5a

View File

@ -5239,6 +5239,10 @@ 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));
/* Traverse only the allowed CPUs */
for_each_cpu_and(i, sched_group_cpus(group), tsk_cpus_allowed(p)) {
if (idle_cpu(i)) {