Messages in this thread | | | Date | Tue, 18 Jun 2013 19:18:20 +0200 | From | Frederic Weisbecker <> | Subject | Re: [PATCH 6/8] sched: task_sched_runtime introduce micro optimization |
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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.
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