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.
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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