Re: PATCH? hrtimer_wakeup: fix a theoretical race wrt rt_mutex_slowlock()

[Oleg Nesterov <oleg@tv-sign.ru>]

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.