On Mon, 2012-09-10 at 22:26 +0530, Srikar Dronamraju wrote:
> > +static bool __yield_to_candidate(struct task_struct *curr, struct task_struct *p)
> > +{
> > + if (!curr->sched_class->yield_to_task)
> > + return false;
> > +
> > + if (curr->sched_class != p->sched_class)
> > + return false;
>
>
> Peter,
>
> Should we also add a check if the runq has a skip buddy (as pointed out
> by Raghu) and return if the skip buddy is already set.
Oh right, I missed that suggestion.. the performance improvement went
from 81% to 139% using this, right?
It might make more sense to keep that separate, outside of this
function, since its not a strict prerequisite.
> >
> > + if (task_running(p_rq, p) || p->state)
> > + return false;
> > +
> > + return true;
> > +}
> > @@ -4323,6 +4340,10 @@ bool __sched yield_to(struct task_struct *p,
> bool preempt)
> > rq = this_rq();
> >
> > again:
> > + /* optimistic test to avoid taking locks */
> > + if (!__yield_to_candidate(curr, p))
> > + goto out_irq;
> > +
So add something like:
/* Optimistic, if we 'raced' with another yield_to(), don't bother */
if (p_rq->cfs_rq->skip)
goto out_irq;
>
>
> > p_rq = task_rq(p);
> > double_rq_lock(rq, p_rq);
>
>
But I do have a question on this optimization though,.. Why do we check
p_rq->cfs_rq->skip and not rq->cfs_rq->skip ?
That is, I'd like to see this thing explained a little better.
Does it go something like: p_rq is the runqueue of the task we'd like to
yield to, rq is our own, they might be the same. If we have a ->skip,
there's nothing we can do about it, OTOH p_rq having a ->skip and
failing the yield_to() simply means us picking the next VCPU thread,
which might be running on an entirely different cpu (rq) and could
succeed?
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