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>
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@ -5239,6 +5239,10 @@ find_idlest_cpu(struct sched_group *group, struct task_struct *p, int this_cpu)
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int shallowest_idle_cpu = -1;
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int shallowest_idle_cpu = -1;
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int i;
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int i;
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/* Check if we have any choice: */
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if (group->group_weight == 1)
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return cpumask_first(sched_group_cpus(group));
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/* Traverse only the allowed CPUs */
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/* Traverse only the allowed CPUs */
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for_each_cpu_and(i, sched_group_cpus(group), tsk_cpus_allowed(p)) {
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for_each_cpu_and(i, sched_group_cpus(group), tsk_cpus_allowed(p)) {
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if (idle_cpu(i)) {
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if (idle_cpu(i)) {
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