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DateMon, 6 Nov 2006 15:09:38 +0300
FromOleg Nesterov <>
SubjectRe: PATCH? hrtimer_wakeup: fix a theoretical race wrt rt_mutex_slowlock()
On 11/05, Linus Torvalds wrote:
> > On Sun, 5 Nov 2006, Steven Rostedt wrote:> > > > 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)> > Yeah, that seems a real bug. You _always_ need to actually do the thing 
> that you wait for _before_ you want it up. That's what all the scheduling 
> primitives depend on - you can't wake people up first, and then set the 
> condition variable.> > So if a rt_mutex depeds on something that is set inside the rq-lock, it 
> needs to get the task rw-lock in order to check it.

No, rt_mutex is fine (I think).

My changelog was very unclean and confusing, I'll try again. What we are
doing is:

	rt_mutex_slowlock:

		task->state = TASK_INTERRUPTIBLE;

		mb();

		if (CONDITION)
			return -ETIMEDOUT;
		schedule();

This is common and correct.

	hrtimer_wakeup:

		CONDITION = 1;			// [1]
		spin_lock(rq->lock);

		task->state = TASK_RUNNING;	// [2]
This needs 'wmb()' between [1] and [2] unless spin_lock() garantees memory
ordering. Of course, rt_mutex can take rq->lock to solve this, but I don't
think it would be right.

Oleg.

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