Re: [RFC][PATCH RT 0/3] RT: Fix trylock deadlock without msleep() hack

[Thomas Gleixner]

On Tue, 8 Sep 2015, Ingo Molnar wrote:
> 
> * Thomas Gleixner <tglx@xxxxxxxxxxxxx> wrote:
> 
> > 3) sched_yield() makes me shudder
> > 
> >    CPU0			CPU1	
> > 
> >    taskA
> >      lock(x->lock)
> > 
> >    preemption
> >    taskC
> > 			taskB
> > 			  lock(y->lock);
> > 			  x = y->x;
> > 			  if (!try_lock(x->lock)) {
> > 			    unlock(y->lock);
> > 			    boost(taskA);
> > 			    sched_yield();  <- returns immediately
> 
> So I'm still struggling with properly parsing the usecase.
> 
> If y->x might become invalid the moment we drop y->lock, what makes
> the 'taskA' use (after we've dropped y->lock) safe? Shouldn't we at
> least also have a task_get(taskA)/task_put(taskA) reference count,
> to make sure the boosted task stays around?

Stevens trylock_and_boost() function makes sure that taskA cannot go
away while doing the boosting. It's a bug in my pseudo code, but that
does not make the issue above going away.
 
> And if we are into getting reference counts, why not solve it at a
> higher level and get a reference count to 'x' to make sure it's safe
> to use? Then we could do:
>
>         lock(y->lock);
> retry:
> 	x = y->x;
>         if (!trylock(x->lock)) {
> 		get_ref(x->count)
>                 unlock(y->lock);
>                 lock(x->lock);
>                 lock(y->lock);
> 		put_ref(x->count);
> 		if (y->x != x) { /* Retry if 'x' got dropped meanwhile */
> 			unlock(x->lock);
> 			goto retry;
> 		}
>         }
> 
> Or so.

In the case of dcache::dentry_kill() we probably do not have to take
refcounts and it might be actually counterproductive to do so. y->x,
i.e. dentry->parent, cannot vanish under us, if I understand the life
time rules correctly.

Aside of that, yes, I was thinking about a similar scheme for
that. I need some more time to grok all the rules there :)

Thanks,

	tglx


 
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