On 11/05, Linus Torvalds wrote:
>
>
> On Sun, 5 Nov 2006, Oleg Nesterov wrote:
> >
> > When task->array != NULL, try_to_wake_up() just goes to "out_running" and sets
> > task->state = TASK_RUNNING.
> >
> > In that case hrtimer_wakeup() does:
> >
> > timeout->task = NULL; <----- [1]
> >
> > spin_lock(runqueues->lock);
> >
> > task->state = TASK_RUNNING; <----- [2]
> >
> > from Documentation/memory-barriers.txt
> >
> > Memory operations that occur before a LOCK operation may appear to
> > happen after it completes.
> >
> > This means that [2] may be completed before [1], and
>
> Yes. On x86 (and x86-64) you'll never see this, because writes are always
> seen in order regardless, and in addition, the spin_lock is actually
> totally serializing anyway. On most other architectures, the spin_lock
> will serialize all the writes too, but it's not guaranteed, so in theory
> you're right. I suspect no actual architecture will do this, but hey,
> when talking memory ordering, safe is a lot better than sorry.
>
> 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.
schedule() will see TASK_INTERRUPTIBLE. hrtimer_wakeup() sets TASK_RUNNING,
rt_mutex_slowlock() does set_current_state(TASK_INTERRUPTIBLE) after that.
schedule() takes runqueue lock, yes, but we are testing timeout->task before.
Oleg.
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