Re: [PATCH 04/19] sched: Prepare for Core-wide rq->lock

[Aubrey Li <aubrey.intel@gmail.com>]

On Fri, Apr 30, 2021 at 4:48 PM Josh Don <joshdon@google.com> wrote:
>
> On Fri, Apr 30, 2021 at 1:20 AM Aubrey Li <aubrey.intel@gmail.com> wrote:
> >
> > On Fri, Apr 30, 2021 at 4:40 AM Josh Don <joshdon@google.com> wrote:
> > >
> > > On Thu, Apr 29, 2021 at 1:03 AM Aubrey Li <aubrey.intel@gmail.com> wrote:
> > > >
> > > > On Thu, Apr 22, 2021 at 8:39 PM Peter Zijlstra <peterz@infradead.org> wrote:
> > > > ----snip----
> > > > > @@ -199,6 +224,25 @@ void raw_spin_rq_unlock(struct rq *rq)
> > > > >         raw_spin_unlock(rq_lockp(rq));
> > > > >  }
> > > > >
> > > > > +#ifdef CONFIG_SMP
> > > > > +/*
> > > > > + * double_rq_lock - safely lock two runqueues
> > > > > + */
> > > > > +void double_rq_lock(struct rq *rq1, struct rq *rq2)
> > > > > +{
> > > > > +       lockdep_assert_irqs_disabled();
> > > > > +
> > > > > +       if (rq1->cpu > rq2->cpu)
> > > >
> > > > It's still a bit hard for me to digest this function, I guess using (rq->cpu)
> > > > can't guarantee the sequence of locking when coresched is enabled.
> > > >
> > > > - cpu1 and cpu7 shares lockA
> > > > - cpu2 and cpu8 shares lockB
> > > >
> > > > double_rq_lock(1,8) leads to lock(A) and lock(B)
> > > > double_rq_lock(7,2) leads to lock(B) and lock(A)
> > > >
> > > > change to below to avoid ABBA?
> > > > +       if (__rq_lockp(rq1) > __rq_lockp(rq2))
> > > >
> > > > Please correct me if I was wrong.
> > >
> > > Great catch Aubrey. This is possibly what is causing the lockups that
> > > Don is seeing.
> > >
> > > The proposed usage of __rq_lockp() is prone to race with sched core
> > > being enabled/disabled.It also won't order properly if we do
> > > double_rq_lock(smt0, smt1) vs double_rq_lock(smt1, smt0), since these
> > > would have equivalent __rq_lockp()
> >
> > If __rq_lockp(smt0) == __rq_lockp(smt1), rq0 and rq1 won't swap,
> > Later only one rq is locked and just returns. I'm not sure how does it not
> > order properly?
>
> If there is a concurrent switch from sched_core enable <-> disable,
> the value of __rq_lockp() will race.
>
> In the version you posted directly above, where we swap rq1 and rq2 if
> __rq_lockp(rq1) > __rqlockp(rq2) rather than comparing the cpu, the
> following can happen:
>
> cpu 1 and cpu 7 share a core lock when coresched is enabled
>
> - schedcore enabled
> - double_lock(7, 1)
> - __rq_lockp compares equal for 7 and 1; no swap is done
> - schedcore disabled; now __rq_lockp returns the per-rq lock
> - lock(__rq_lockp(7)) => lock(7)
> - lock(__rq_lockp(1)) => lock(1)
>
> Then we can also have
>
> - schedcore disabled
> - double_lock(1, 7)
> - __rq_lock(1) < rq_lock(7), so no swap
> - lock(__rqlockp(1)) => lock(1)
> - lock(__rq_lockp(7)) => lock(7)
>
> So we have in the first 7->1 and in the second 1->7
>
> >
> > .> I'd propose an alternative but similar idea: order by core, then break ties
> > > by ordering on cpu.
> > >
> > > +#ifdef CONFIG_SCHED_CORE
> > > +       if (rq1->core->cpu > rq2->core->cpu)
> > > +               swap(rq1, rq2);
> > > +       else if (rq1->core->cpu == rq2->core->cpu && rq1->cpu > rq2->cpu)
> > > +               swap(rq1, rq2);
> >
> > That is, why the "else if" branch is needed?
>
> Ensuring that core siblings always take their locks in the same order
> if coresched is disabled.
>

Both this and above make sense to me. Thanks for the great elaboration, Josh!

Thanks,
-Aubrey