Re: [PATCH RFC 5/5] sched/fair: remove impossible condition from find_idlest_group_cpu

[Vincent Guittot]

On 28 October 2017 at 11:59, Joel Fernandes <joelaf@xxxxxxxxxx> wrote:
> find_idlest_group_cpu goes through CPUs of a group previous selected by
> find_idlest_group. find_idlest_group returns NULL if the local group is the
> selected one and doesn't execute find_idlest_group_cpu if the group to which
> 'cpu' belongs to is chosen. So we're always guaranteed to call
> find_idlest_group_cpu with a group to which cpu is non-local. This makes one of

Is this still true in case of overlapping topology ?

> the conditions in find_idlest_group_cpu an impossible one, which we can get rid
> off.
>
> Cc: Ingo Molnar <mingo@xxxxxxxxxx>
> Cc: Peter Zijlstra <peterz@xxxxxxxxxxxxx>
> Cc: Brendan Jackman <brendan.jackman@xxxxxxx>
> Cc: Dietmar <dietmar.eggemann@xxxxxxx>
> Signed-off-by: Joel Fernandes <joelaf@xxxxxxxxxx>
> ---
>  kernel/sched/fair.c | 2 +-
>  1 file changed, 1 insertion(+), 1 deletion(-)
>
> diff --git a/kernel/sched/fair.c b/kernel/sched/fair.c
> index 5c49fdb4c508..740602ce799f 100644
> --- a/kernel/sched/fair.c
> +++ b/kernel/sched/fair.c
> @@ -5922,7 +5922,7 @@ find_idlest_group_cpu(struct sched_group *group, struct task_struct *p, int this
>                         }
>                 } else if (shallowest_idle_cpu == -1) {
>                         load = weighted_cpuload(cpu_rq(i));
> -                       if (load < min_load || (load == min_load && i == this_cpu)) {
> +                       if (load < min_load) {
>                                 min_load = load;
>                                 least_loaded_cpu = i;
>                         }
> --
> 2.15.0.rc2.357.g7e34df9404-goog
>