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Date:	Sun, 5 Nov 2006 17:53:32 -0500 (EST)
From:	Steven Rostedt <rostedt@...dmis.org>
To:	Linus Torvalds <torvalds@...l.org>
cc:	Oleg Nesterov <oleg@...sign.ru>,
	Thomas Gleixner <tglx@...utronix.de>,
	Andrew Morton <akpm@...l.org>, linux-kernel@...r.kernel.org
Subject: Re: PATCH? hrtimer_wakeup: fix a theoretical race wrt rt_mutex_slowlock()


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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