Re: [PATCH 6/8] sched: task_sched_runtime introduce micro optimization

[Frederic Weisbecker <fweisbec@gmail.com>]

On Tue, Jun 18, 2013 at 11:17:41AM -0400, KOSAKI Motohiro wrote:
> >> +#ifdef CONFIG_64BIT
> >> +     /*
> >> +      * 64-bit doesn't need locks to atomically read a 64bit value. So we
> >> +      * have two optimization chances, 1) when caller doesn't need
> >> +      * delta_exec and 2) when the task's delta_exec is 0. The former is
> >> +      * obvious. The latter is complicated. reading ->on_cpu is racy, but
> >> +      * this is ok. If we race with it leaving cpu, we'll take a lock. So
> >> +      * we're correct. If we race with it entering cpu, unaccounted time
> >> +      * is 0. This is indistinguishable from the read occurring a few
> >> +      * cycles earlier.
> >> +      */
> >> +     if (!add_delta || !p->on_cpu)
> >> +             return p->se.sum_exec_runtime;
> >
> > I'm not sure this is correct from an smp ordering POV. p->on_cpu may appear
> > to be 0 whereas the task is actually running for a while and p->se.sum_exec_runtime
> > can then be past the actual value on the remote CPU.
> 
> Quate form Paul's last e-mail
> 
> >Stronger:
> >
> >+#ifdef CONFIG_64BIT
> >+       if (!p->on_cpu)
> >+               return p->se.sum_exec_runtime;
> >+#endif
> >
> >[ Or !p->on_cpu || !add_delta ].
> >
> >We can take the racy read versus p->on_cpu since:
> >  If we race with it leaving cpu: we take lock, we're correct
> >  If we race with it entering cpu: unaccounted time ---> 0, this is
> >indistinguishable from the read occurring a few cycles earlier.

Yeah, my worry was more about both p->on_cpu and p->se.sum_exec_runtime being
stale for too long. How much time can happen in the worst case before CPU X sees
the updates done by a CPU Y under rq(Y)->lock considering that CPU X doesn't take rq(Y)
to read that update? I guess it depends on the hardware, locking and ordering
that happened before.

Bah it probably doesn't matter in practice.

Thanks.