Re: [PATCH] de_thread: Move notify_count write under lock

[Kirill Tkhai <ktkhai@parallels.com>]

В Чт, 05/02/2015 в 14:38 +0100, Oleg Nesterov пишет:
> On 02/05, Kirill Tkhai wrote:
> >
> > The write operation may be reordered with the setting of group_exit_task.
> > If so, this fires in exit_notify().
> 
> How?
> 
> OK, yes, "sig->notify_count = -1" can be reordered with the last unlock,
> but we do not care?
> 
> group_exit_task + notify_count is only checked under the same lock, and
> "notify_count = -1" can't happen until de_thread() sees it is zero.
> 
> Could you explain why this is bad in more details?

Can't exit_notify() see tsk->signal->notify_count == -1 before
tsk->signal->group_exit_task?

As I see in Documentation/memory-barriers.txt:

	RELEASE operation implication:
		Memory operations issued after the RELEASE may be completed before the
		RELEASE operation has completed.


> 
> > --- a/fs/exec.c
> > +++ b/fs/exec.c
> > @@ -920,10 +920,16 @@ static int de_thread(struct task_struct *tsk)
> >  	if (!thread_group_leader(tsk)) {
> >  		struct task_struct *leader = tsk->group_leader;
> >
> > -		sig->notify_count = -1;	/* for exit_notify() */
> >  		for (;;) {
> >  			threadgroup_change_begin(tsk);
> >  			write_lock_irq(&tasklist_lock);
> > +			/*
> > +			 * We could set it once outside the for() cycle, but
> > +			 * this requires to use SMP barriers there and in
> > +			 * exit_notify(), because the write operation may
> > +			 * be reordered with the setting of group_exit_task.
> > +			 */
> > +			sig->notify_count = -1;	/* for exit_notify() */
> >  			if (likely(leader->exit_state))
> >  				break;
> >  			__set_current_state(TASK_KILLABLE);
> 
> Perhaps something like this makes sense anyway to make the code more
> clear, but in this case I'd suggest to set ->notify_count after we
> check ->exit_state. And without the (afaics!) misleading comment...
> 
> Or I missed something?
> 
> Oleg.
>