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

[Joel Fernandes]

Hi Vincent,

On Mon, Oct 30, 2017 at 9:00 AM, Vincent Guittot
<vincent.guittot@xxxxxxxxxx> wrote:
> 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 ?

Yes, I believe so. As per the code, find_idlest_group will only return
a group to which this_cpu doesn't belong to. So incase an overlapping
group was returned by find_idlest_group (instead of NULL), then none
of the groups (among the set of overlapping groups) is local to
this_cpu. Incase NULL is returned, then find_idlest_group_cpu doesn't
execute at all.

Do you agree?

thanks,

- Joel

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