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sched/fair: Fix usage of find_idlest_group() when no groups are allowed

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When 'p' is not allowed on any of the CPUs in the sched_domain, we
currently return NULL from find_idlest_group(), and pointlessly
continue the search on lower sched_domain levels (where 'p' is also not
allowed) before returning prev_cpu regardless (as we have not updated
new_cpu).

Add an explicit check for this case, and add a comment to
find_idlest_group(). Now when find_idlest_group() returns NULL, it always
means that the local group is allowed and idlest.

Signed-off-by: Brendan Jackman <brendan.jackman@arm.com>
Signed-off-by: Peter Zijlstra (Intel) <peterz@infradead.org>
Reviewed-by: Vincent Guittot <vincent.guittot@linaro.org>
Reviewed-by: Josef Bacik <jbacik@fb.com>
Cc: Dietmar Eggemann <dietmar.eggemann@arm.com>
Cc: Josef Bacik <josef@toxicpanda.com>
Cc: Linus Torvalds <torvalds@linux-foundation.org>
Cc: Mike Galbraith <efault@gmx.de>
Cc: Morten Rasmussen <morten.rasmussen@arm.com>
Cc: Peter Zijlstra <peterz@infradead.org>
Cc: Thomas Gleixner <tglx@linutronix.de>
Link: http://lkml.kernel.org/r/20171005114516.18617-5-brendan.jackman@arm.com
Signed-off-by: Ingo Molnar <mingo@kernel.org>
This commit is contained in:
Brendan Jackman 2017-10-05 12:45:15 +01:00 committed by Ingo Molnar
parent 0d10ab952e
commit 6fee85ccbc
1 changed files with 5 additions and 0 deletions
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5
kernel/sched/fair.c
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@ -5730,6 +5730,8 @@ static unsigned long capacity_spare_wake(int cpu, struct task_struct *p)
/*
* find_idlest_group finds and returns the least busy CPU group within the
* domain.
*
* Assumes p is allowed on at least one CPU in sd.
*/
static struct sched_group *
find_idlest_group(struct sched_domain *sd, struct task_struct *p,
@ -5917,6 +5919,9 @@ static inline int find_idlest_cpu(struct sched_domain *sd, struct task_struct *p
{
int new_cpu = prev_cpu;
if (!cpumask_intersects(sched_domain_span(sd), &p->cpus_allowed))
return prev_cpu;
while (sd) {
struct sched_group *group;
struct sched_domain *tmp;