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

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 <[email protected]>
Signed-off-by: Peter Zijlstra (Intel) <[email protected]>
Reviewed-by: Vincent Guittot <[email protected]>
Reviewed-by: Josef Bacik <[email protected]>
Cc: Dietmar Eggemann <[email protected]>
Cc: Josef Bacik <[email protected]>
Cc: Linus Torvalds <[email protected]>
Cc: Mike Galbraith <[email protected]>
Cc: Morten Rasmussen <[email protected]>
Cc: Peter Zijlstra <[email protected]>
Cc: Thomas Gleixner <[email protected]>
Link: http://lkml.kernel.org/r/[email protected]
Signed-off-by: Ingo Molnar <[email protected]>

1 files changed, 5 insertions, 0 deletions

diff --git a/kernel/sched/fair.c b/kernel/sched/fair.c
index cca1835efa7b..ed80d6bd76c8 100644
--- a/kernel/sched/fair.c
+++ b/kernel/sched/fair.c
@@ -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;