Re: [PATCH v10 03/20] timers: Move marking timer bases idle into tick_nohz_stop_tick()

Frederic Weisbecker
2024-01-17T16:02:30Z

Le Mon, Jan 15, 2024 at 03:37:26PM +0100, Anna-Maria Behnsen a écrit :
> @@ -889,12 +884,41 @@ static ktime_t tick_nohz_next_event(struct tick_sched *ts, int cpu)
> static void tick_nohz_stop_tick(struct tick_sched *ts, int cpu)
> {
> struct clock_event_device *dev = __this_cpu_read(tick_cpu_device.evtdev);
> + unsigned long basejiff = ts->last_jiffies;
> u64 basemono = ts->timer_expires_base;
> - u64 expires = ts->timer_expires;
> + bool timer_idle;
> + u64 expires;
>
> /* Make sure we won't be trying to stop it twice in a row. */
> ts->timer_expires_base = 0;
>
> + /*
> + * Now the tick should be stopped definitely - so the timer base needs
> + * to be marked idle as well to not miss a newly queued timer.
> + */
> + expires = timer_base_try_to_set_idle(basejiff, basemono, &timer_idle);
> + if (!timer_idle) {
> + /*
> + * Do not clear tick_stopped here when it was already set - it

Can that really happen? Looking at __get_next_timer_interrupt(), you're making a
behavioural change: if base->is_idle was previously set and the next timer is
now below/equal a jiffy, base->is_idle is not going to be cleared by
__get_next_timer_interrupt().

Therefore you shouldn't observe ts->tick_stopped && !timer_idle

But I'm assuming that behavioural change wasn't intended?

> + * will be retained on the next idle iteration when the tick
> + * expired earlier than expected.

I'm a bit confused by this sentence.

> + */
> + expires = basemono + TICK_NSEC;

Do you need this line?

> @@ -1147,11 +1175,6 @@ void tick_nohz_idle_stop_tick(void)
> void tick_nohz_idle_retain_tick(void)
> {
> tick_nohz_retain_tick(this_cpu_ptr(&tick_cpu_sched));

Looks like the content of tick_nohz_retain_tick() can move here now.

> - /*
> - * Undo the effect of get_next_timer_interrupt() called from
> - * tick_nohz_next_event().
> - */
> - timer_clear_idle();
> }

Thanks.