Re: [RFC] Deadlock via recursive wakeup via RCU with threadirqs

Paul E. McKenney
2019-06-28T12:27:45Z

On Fri, Jun 28, 2019 at 07:40:45PM +0900, Byungchul Park wrote:
> On Fri, Jun 28, 2019 at 04:31:38PM +0900, Byungchul Park wrote:
> > On Thu, Jun 27, 2019 at 01:36:12PM -0700, Paul E. McKenney wrote:
> > > On Thu, Jun 27, 2019 at 03:17:27PM -0500, Scott Wood wrote:
> > > > On Thu, 2019-06-27 at 11:41 -0700, Paul E. McKenney wrote:
> > > > > On Thu, Jun 27, 2019 at 02:16:38PM -0400, Joel Fernandes wrote:
> > > > > >
> > > > > > I think the fix should be to prevent the wake-up not based on whether we
> > > > > > are
> > > > > > in hard/soft-interrupt mode but that we are doing the rcu_read_unlock()
> > > > > > from
> > > > > > a scheduler path (if we can detect that)
> > > > >
> > > > > Or just don't do the wakeup at all, if it comes to that. I don't know
> > > > > of any way to determine whether rcu_read_unlock() is being called from
> > > > > the scheduler, but it has been some time since I asked Peter Zijlstra
> > > > > about that.
> > > > >
> > > > > Of course, unconditionally refusing to do the wakeup might not be happy
> > > > > thing for NO_HZ_FULL kernels that don't implement IRQ work.
> > > >
> > > > Couldn't smp_send_reschedule() be used instead?
> > >
> > > Good point. If current -rcu doesn't fix things for Sebastian's case,
> > > that would be well worth looking at. But there must be some reason
> > > why Peter Zijlstra didn't suggest it when he instead suggested using
> > > the IRQ work approach.
> > >
> > > Peter, thoughts?
> >
> > Hello,
> >
> > Isn't the following scenario possible?
> >
> > The original code
> > -----------------
> > rcu_read_lock();
> > ...
> > /* Experdite */
> > WRITE_ONCE(t->rcu_read_unlock_special.b.exp_hint, true);
> > ...
> > __rcu_read_unlock();
> > if (unlikely(READ_ONCE(t->rcu_read_unlock_special.s)))
> > rcu_read_unlock_special(t);
> > WRITE_ONCE(t->rcu_read_unlock_special.b.exp_hint, false);
> > rcu_preempt_deferred_qs_irqrestore(t, flags);
> > barrier(); /* ->rcu_read_unlock_special load before assign */
> > t->rcu_read_lock_nesting = 0;
> >
> > The reordered code by machine
> > -----------------------------
> > rcu_read_lock();
> > ...
> > /* Experdite */
> > WRITE_ONCE(t->rcu_read_unlock_special.b.exp_hint, true);
> > ...
> > __rcu_read_unlock();
> > if (unlikely(READ_ONCE(t->rcu_read_unlock_special.s)))
> > rcu_read_unlock_special(t);
> > t->rcu_read_lock_nesting = 0; <--- LOOK AT THIS!!!
> > WRITE_ONCE(t->rcu_read_unlock_special.b.exp_hint, false);
> > rcu_preempt_deferred_qs_irqrestore(t, flags);
> > barrier(); /* ->rcu_read_unlock_special load before assign */
> >
> > An interrupt happens
> > --------------------
> > rcu_read_lock();
> > ...
> > /* Experdite */
> > WRITE_ONCE(t->rcu_read_unlock_special.b.exp_hint, true);
> > ...
> > __rcu_read_unlock();
> > if (unlikely(READ_ONCE(t->rcu_read_unlock_special.s)))
> > rcu_read_unlock_special(t);
> > t->rcu_read_lock_nesting = 0; <--- LOOK AT THIS!!!
> > <--- Handle an (any) irq
> > rcu_read_lock();
> > /* This call should be skipped */
> > rcu_read_unlock_special(t);
>
> Wait.. I got a little bit confused on recordering.
>
> This 'STORE rcu_read_lock_nesting = 0' can happen before
> 'STORE rcu_read_unlock_special.b.exp_hint = false' regardless of the
> order a compiler generated to by the barrier(), because anyway they
> are independent so it's within an arch's right.
>
> Then.. is this scenario possible? Or all archs properly deal with
> interrupts across this kind of reordering?

Interrupts are "exact" in that they happen between a pair of consecutive
instructions from the viewpoint of the CPU taking the interrupt. And
again, these fields are local to their CPU/task, give or take debug code.

Thanx, Paul

> Thanks,
> Byungchul
>
> > WRITE_ONCE(t->rcu_read_unlock_special.b.exp_hint, false);
> > rcu_preempt_deferred_qs_irqrestore(t, flags);
> > barrier(); /* ->rcu_read_unlock_special load before assign */
> >
> > We don't have to handle the special thing twice like this which is one
> > reason to cause the problem even though another problem is of course to
> > call ttwu w/o being aware it's within a context holding pi lock.
> >
> > Apart from the discussion about how to avoid ttwu in an improper
> > condition, I think the following is necessary. I may have something
> > missing. It would be appreciated if you let me know in case I'm wrong.
> >
> > Anyway, logically I think we should prevent reordering between
> > t->rcu_read_lock_nesting and t->rcu_read_unlock_special.b.exp_hint not
> > only by compiler but also by machine like the below.
> >
> > Do I miss something?
> >
> > Thanks,
> > Byungchul
> >
> > ---8<---
> > diff --git a/kernel/rcu/tree_plugin.h b/kernel/rcu/tree_plugin.h
> > index 3c8444e..9b137f1 100644
> > --- a/kernel/rcu/tree_plugin.h
> > +++ b/kernel/rcu/tree_plugin.h
> > @@ -412,7 +412,13 @@ void __rcu_read_unlock(void)
> > barrier(); /* assign before ->rcu_read_unlock_special load */
> > if (unlikely(READ_ONCE(t->rcu_read_unlock_special.s)))
> > rcu_read_unlock_special(t);
> > - barrier(); /* ->rcu_read_unlock_special load before assign */
> > + /*
> > + * Prevent reordering between clearing
> > + * t->rcu_reak_unlock_special in
> > + * rcu_read_unlock_special() and the following
> > + * assignment to t->rcu_read_lock_nesting.
> > + */
> > + smp_wmb();
> > t->rcu_read_lock_nesting = 0;
> > }
> > if (IS_ENABLED(CONFIG_PROVE_LOCKING)) {
> >
> >
>