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sched/topology: Simplify sched_group_mask() usage

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While writing the comments, it occurred to me that:

  sg_cpus & sg_mask == sg_mask

at least conceptually; the !overlap case sets the all 1s mask. If we
correct that we can simplify things and directly use sg_mask.

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: linux-kernel@vger.kernel.org
Signed-off-by: Ingo Molnar <mingo@kernel.org>
This commit is contained in:
Peter Zijlstra
2017-05-01 08:51:05 +02:00
committed by Ingo Molnar
parent 0c0e776a9b
commit af218122b1
2 changed files with 6 additions and 6 deletions
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5
kernel/sched/topology.c
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@@ -85,7 +85,8 @@ static int sched_domain_debug_one(struct sched_domain *sd, int cpu, int level,
group->sgc->id,
cpumask_pr_args(sched_group_cpus(group)));
if ((sd->flags & SD_OVERLAP) && !cpumask_full(sched_group_mask(group))) {
if ((sd->flags & SD_OVERLAP) &&
!cpumask_equal(sched_group_mask(group), sched_group_cpus(group))) {
printk(KERN_CONT " mask=%*pbl",
cpumask_pr_args(sched_group_mask(group)));
}
@@ -505,7 +506,7 @@ enum s_alloc {
*/
int group_balance_cpu(struct sched_group *sg)
{
return cpumask_first_and(sched_group_cpus(sg), sched_group_mask(sg));
return cpumask_first(sched_group_mask(sg));
}