Re: [PATCH v2] CPU hotplug: active_writer not woken up in some cases - deadlock

[David Hildenbrand]

> The compiler is within its rights to optimize the active_writer local
> variable out of existence, thus re-introducing the possible race with
> the writer that can pass a NULL pointer to wake_up_process().  So you
> really need the ACCESS_ONCE() on the read from cpu_hotplug.active_writer.
> Please see http://lwn.net/Articles/508991/ for more information why
> this is absolutely required.

You're absolutely right, saw your reply on the other patch just after I sent
this version ...

So if you agree with the change below, I'll send an updated version!

> 
> > +		if (unlikely(active_writer))
> > +			wake_up_process(active_writer);
> >  		cpuhp_lock_release();
> >  		return;
> >  	}
> > @@ -161,15 +167,17 @@ void cpu_hotplug_begin(void)
> >  	cpuhp_lock_acquire();
> >  	for (;;) {
> >  		mutex_lock(&cpu_hotplug.lock);
> > +		__set_current_state(TASK_UNINTERRUPTIBLE);
> 
> You lost me on this one.  How does this help?
> 
> 							Thanx, Paul

Imagine e.g. the following (simplified) scenario:

CPU1                               CPU2
----------------------------------------------------------------------------
!mutex_trylock(&cpu_hotplug.lock) |
                                  | cpu_hotplug.puts_pending == 0
cpu_hotplug.puts_pending++;       |
                                  | cpu_hotplug.refcount != 0
wake_up_process(active_writer)
                                  | __set_current_state(TASK_UNINTERRUPTIBLE);
                                  | schedule();
                                  | /* will never be woken up */

Therefore we have to move the condition check inside the 
  __set_current_state(TASK_UNINTERRUPTIBLE) -> schedule();
section to not miss any wake ups when the condition is satisfied.

So wake_up_process() will either see TASK_RUNNING and do nothing or see
TASK_UNINTERRUPTIBLE and set it to TASK_RUNNING, so schedule() will in
fact be woken up again.


Thanks a lot!

David

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