On 02/25, Rafael J. Wysocki wrote:
>
> On Sunday, 25 February 2007 16:40, Aneesh Kumar wrote:
> > On 2/25/07, Aneesh Kumar <[EMAIL PROTECTED]> wrote:
> > > On 2/25/07, Rafael J. Wysocki <[EMAIL PROTECTED]> wrote:
> > > > On Sunday, 25 February 2007 15:33, Aneesh Kumar wrote:
> > > > > On 2/25/07, Rafael J. Wysocki <[EMAIL PROTECTED]> wrote:
> [--snip--]
> >
> > Thinking about this i guess we have a problem with the above approach
> > i outlined. if we have one task that is waiting on the event and more
> > than one that can generate the event then the above logic would not
> > work. Also with cases other than vfork; logic of tracking the waiting
> > task gets complex. I guess what we have right now is better.
>
> I assume by "righ now" you mean the latest version of my patch. ;-)
>
> Still, having pondered the Pavel's suggestion for a while I think it's doable
> without the addtitional process flag. Patch below.
Probably I missed something, (I didn't see this patch and I missed the
start of discussion), but I can't understand this patch.
> +__wait_for_completion(struct completion *x, int freezable)
> {
> might_sleep();
>
> @@ -3817,6 +3818,9 @@ void fastcall __sched wait_for_completio
> __set_current_state(TASK_UNINTERRUPTIBLE);
> spin_unlock_irq(&x->wait.lock);
> schedule();
> + if (freezable)
> + try_to_freeze();
> +
> spin_lock_irq(&x->wait.lock);
> } while (!x->done);
> __remove_wait_queue(&x->wait, &wait);
> @@ -3824,7 +3828,7 @@ void fastcall __sched wait_for_completio
> x->done--;
> spin_unlock_irq(&x->wait.lock);
> }
>
> ..........
>
> @@ -48,6 +48,9 @@ void refrigerator(void)
> task_unlock(current);
> return;
> }
> + if (current->vfork_done)
> + wake_up_process(current->parent);
> +
What if current->parent doesn't have TIF_FREEZE yet? ->parent will schedule()
again, child goes to refrigerator. Now, how can we freeze the ->parent?
Oleg.
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