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SubjectRe: [BUG] TASK_DEAD task is able to be woken up in special condition
On 01/24, Peter Zijlstra wrote:
>
> On Tue, 2012-01-24 at 11:19 +0100, Peter Zijlstra wrote:
> > On Wed, 2012-01-18 at 15:20 +0100, Oleg Nesterov wrote:
> > > do_exit() is different because it can not handle the spurious wakeup.
> > > Well, may be we can? we can simply do
> > >
> > > for (;;) {
> > > tsk->state = TASK_DEAD;
> > > schedule();
> > > }
> > >
> > > __schedule() can't race with ttwu() once it takes rq->lock. If the
> > > exiting task is deactivated, finish_task_switch() will see EXIT_DEAD.
> >
> > TASK_DEAD, right?

Yes, but... I simply can't understand what I was thinking about.
And probably I missed something again, but I think this can't work.

Afaics, this can only help to prevent the race with ttwu_remote()
doing ttwu_do_wakeup() under rq->lock.

But we still can race with the !p->on_rq case which sets TASK_WAKING.
It can do this after finish_task_switch() observes TASK_DEAD and does
put_task_struct().

> I think Yasunori-San's patch isn't
> sufficient, note how the p->state = TASK_RUNNING in ttwu_do_wakeup() can
> happen outside of p->pi_lock when the task gets queued on a remote cpu.

Hmm, really? I am not sure, but I do not trust myself.

To simplify, you mean that

mb();
unlock_wait(pi_lock);

tsk->state = TASK_DEAD;

can change ->state from TASK_WAKING to TASK_DEAD, right? Is this really
possible? ttwu() ensures p->on_rq == F in this case.

Oleg.



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