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SubjectRe: [PATCH v2 1/3] sched: Stop nohz stats when decayed
On Tue, Feb 06, 2018 at 08:23:05PM +0100, Vincent Guittot wrote:

> @@ -9207,13 +9231,15 @@ void nohz_balance_enter_idle(int cpu)
> if (!housekeeping_cpu(cpu, HK_FLAG_SCHED))
> return;
>
> + rq->has_blocked_load = 1;
> +
> if (rq->nohz_tick_stopped)
> - return;

this case is difficult... needs thinking

> + goto out;
>
> /*
> * If we're a completely isolated CPU, we don't play.
> */
> - if (on_null_domain(cpu_rq(cpu)))
> + if (on_null_domain(rq))
> return;
>
> rq->nohz_tick_stopped = 1;
> @@ -9222,6 +9248,13 @@ void nohz_balance_enter_idle(int cpu)
> atomic_inc(&nohz.nr_cpus);
>
> set_cpu_sd_state_idle(cpu);

/*
* Ensures that if nohz_idle_balance() fails to observe our
* @idle_cpus_mask store, it must observe the @has_blocked
* store.
*/
smp_mb__after_atomic();

> +
> +out:
> + /*
> + * Each time a cpu enter idle, we assume that it has blocked load and
> + * enable the periodic update of the load of idle cpus
> + */
> + WRITE_ONCE(nohz.has_blocked, 1);
> }



> @@ -9374,6 +9407,16 @@ static bool nohz_idle_balance(struct rq *this_rq, enum cpu_idle_type idle)
>
> SCHED_WARN_ON((flags & NOHZ_KICK_MASK) == NOHZ_BALANCE_KICK);
>
> + /*
> + * We assume there will be no idle load after this update and clear
> + * the has_blocked flag. If a cpu enters idle in the mean time, it will
> + * set the has_blocked flag and trig another update of idle load.
> + * Because a cpu that becomes idle, is added to idle_cpus_mask before
> + * setting the flag, we are sure to not clear the state and not
> + * check the load of an idle cpu.
> + */
> + WRITE_ONCE(nohz.has_blocked, 0);

/*
* Ensures that if we miss the CPU, we must see the has_blocked
* store from nohz_balance_enter_idle().
*/
smp_mb();

> for_each_cpu(balance_cpu, nohz.idle_cpus_mask) {
> if (balance_cpu == this_cpu || !idle_cpu(balance_cpu))
> continue;


I _think_, but my brain isn't quite willing to turn on today.

Without this ordering I think it would be possible to loose has_blocked
and not observe the CPU either.

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