On Thu, Aug 20, 2015 at 08:35:16PM +0900, Byungchul Park wrote:
> On Thu, Aug 20, 2015 at 08:22:00PM +0900, byungchul.p...@lge.com wrote:
> > + /*
> > + * If it's !queued, then only when the task is sleeping it has a
> > + * non-normalized vruntime, that is, when the task is being migrated
> > + * it has a normailized vruntime.
> > + */
>
> i tried to change your XXX comment. i think it can be explaned like this.
> don't you think so? i want to hear any opinions about this.
>
> > + if (p->state == TASK_RUNNING)
> > + return true;
--- a/kernel/sched/fair.c
+++ b/kernel/sched/fair.c
@@ -7943,11 +7943,10 @@ static inline bool vruntime_normalized(s
return true;
/*
- * If it's !queued, then only when the task is sleeping it has a
- * non-normalized vruntime, that is, when the task is being migrated
- * it has a normalized vruntime.
+ * If it's !queued, sleeping tasks have a normalized vruntime,
+ * see dequeue_entity().
*/
- if (p->state == TASK_RUNNING)
+ if (!p->se.on_rq)
return true;
return false;
Does that make sense?
I think using p->state for this is fragile, as we could be racy with any
random blocking primitive that does set_current_state() _before_
actually calling into the scheduler.
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