On Thursday, 2 August 2007 20:40, Oleg Nesterov wrote:
> On 08/02, Rafael J. Wysocki wrote:
> >
> > @@ -171,6 +186,10 @@ static int try_to_freeze_tasks(int freez
> >
> > end_time = jiffies + TIMEOUT;
> > do {
> > + DEFINE_WAIT(wait);
> > +
> > + add_wait_queue(&refrigerator_waitq, &wait);
>
> Hmm. In that case I'd sugest to use prepare_to_wait(). This means that
> multiple wakeups from refrigerator() won't do unnecessary work,
I'm not sure what you mean.
Do you mean that if we are TASK_UNINTERRUPTIBLE, then the first wake up
should remove us from the queue?
> and
>
> > +
> > todo = 0;
> > read_lock(&tasklist_lock);
> > do_each_thread(g, p) {
> > @@ -189,7 +208,12 @@ static int try_to_freeze_tasks(int freez
> > todo++;
> > } while_each_thread(g, p);
> > read_unlock(&tasklist_lock);
> > - yield(); /* Yield is okay here */
> > +
> > + set_current_state(TASK_UNINTERRUPTIBLE);
> > + if (todo && !list_empty_careful(&wait.task_list))
> > + schedule_timeout(WAIT_TIME);
>
> we don't need to check list_empty_careful() before schedule, prepare_to_wait()
> sets TASK_UNINTERRUPTIBLE under wait_queue_head_t->lock.
Yes.
> Still, I personally agree with Pavel. Perhaps it is better to just replace
> yield() with schedule_timeout(a_bit).
Hmm, I think that we shouldn't wait if that's not necessary.
Greetings,
Rafael
--
"Premature optimization is the root of all evil." - Donald Knuth
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