On 11/05, Steven Rostedt wrote:
>
> On Sun, 5 Nov 2006, Linus Torvalds wrote:
>
> >
> > That said, since "task->state" in only tested _inside_ the runqueue lock,
> > there is no race that I can see. Since we've gotten the runqueue lock in
> > order to even check task-state, the processor that _sets_ task state must
> > not only have done the "spin_lock()", it must also have done the
> > "spin_unlock()", and _that_ will not allow either the timeout or the task
> > state to haev leaked out from under it (because that would imply that the
> > critical region leaked out too).
> >
> > So I don't think the race exists anyway - the schedule() will return
> > immediately (because it will see TASK_RUNNING), and we'll just retry.
> >
>
> This whole situation is very theoretical, but I think this actually can
> happen *theoretically*.
>
>
> OK, the spin_lock doesn't do any serialization, but the unlock does. But
> the problem can happen before the unlock. Because of the loop.
Yes, yes, thanks.
( Actually, this was more "is my understanding correct?" than a patch )
Thanks!
> CPU 1 CPU 2
>
> task_rq_lock()
>
> p->state = TASK_RUNNING;
>
>
> (from bottom of for loop)
> set_current_state(TASK_INTERRUPTIBLE);
>
> for (;;) { (looping)
>
> if (timeout && !timeout->task)
>
>
> (now CPU implements)
> t->task = NULL
>
> task_rq_unlock();
>
> schedule() (with state == TASK_INTERRUPTIBLE)
>
>
> Again, this is very theoretical, and I don't even think that this can
> happen if you tried to make it. But I guess if hardware were to change in
> the future with the same rules that we have today with barriers, that this
> can be a race.
>
> -- Steve
>
>
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