Messages in this thread | | | Date | Wed, 25 Jan 2012 16:45:47 +0100 | From | Oleg Nesterov <> | Subject | Re: [BUG] TASK_DEAD task is able to be woken up in special condition |
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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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