Re: [PATCH 2/2] sched/fair: Randomize equally shallow slow-path candidates

[Shubhang <sh@gentwo.org>]

Hi Christian,

On Wed, 16 Sep 2026, Christian Loehle wrote:

> +			} else if (!idle || idle->exit_latency == min_exit_latency) {
> +				nr_candidates++;
> +				if (nr_candidates == 1 ||
> +				    !reciprocal_scale(sched_rng(), nr_candidates))
> +					shallowest_idle_cpu = i;

available_idle_cpu(i) ensures that this is an idle CPU, but !idle 
means that no active cpuidle state, meaning no exit latency is 
available for comparison.

[PATCH 1/2] treats such a CPU as a fallback i.e. it is selected only when 
no idle candidate has been found yet. Here it becomes an equal reservoir 
candidate, even after selecting a CPU with the minimum known exit latency. 
That is, it is added to the random selection pool and can replace 
shallowest_idle_cpu.

Is that intentional ? If not, should reservoir sampling be limited to
candidates with `idle->exit_latency == min_exit_latency`, while 
retaining the first !idle CPU only as the fallback?

Regards,
Shubhang Kaushik