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SubjectRe: [PATCH v2] sched/fair: Remove the cost of a redundant cpumask_next_wrap in select_idle_cpu
On Thu, Nov 25, 2021 at 12:49 AM Barry Song <21cnbao@gmail.com> wrote:
>
> On Thu, Nov 25, 2021 at 12:13 AM Peter Zijlstra <peterz@infradead.org> wrote:
> >
> > On Wed, Nov 24, 2021 at 05:15:46PM +0800, Barry Song wrote:
> > > diff --git a/kernel/sched/fair.c b/kernel/sched/fair.c
> > > index 6e476f6..8cd23f1 100644
> > > --- a/kernel/sched/fair.c
> > > +++ b/kernel/sched/fair.c
> > > @@ -6278,6 +6278,7 @@ static int select_idle_cpu(struct task_struct *p, struct sched_domain *sd, bool
> > > time = cpu_clock(this);
> > > }
> > >
> > > + --nr;
> > > for_each_cpu_wrap(cpu, cpus, target + 1) {
> > > if (has_idle_core) {
> > > i = select_idle_core(p, cpu, cpus, &idle_cpu);
> > > @@ -6285,11 +6286,11 @@ static int select_idle_cpu(struct task_struct *p, struct sched_domain *sd, bool
> > > return i;
> > >
> > > } else {
> > > - if (!--nr)
> > > - return -1;
> > > idle_cpu = __select_idle_cpu(cpu, p);
> > > if ((unsigned int)idle_cpu < nr_cpumask_bits)
> > > break;
> > > + if (!--nr)
> > > + return -1;
> > > }
> > > }
> >
> > This way nr can never be 1 for a single iteration -- it current isn't,
> > but that's besides the point.
>
> Yep. nr=1 seems to be a corner case.
> if nr=1, the original code will return -1 directly without scanning
> any cpu. but the new code will scan
> one time because I haven't checked if(!--nr) and return before
> for_each_cpu_wrap(). so this changes
> the behaviour for this corner case.
>
> but if i change "--nr" to "nr--", this new code will scan nr times
> rather than nr-1, this changes the behaviour
> for all cases besides nr!=1. The original code was looping nr times
> but scanning nr-1 times only.
>
> so you prefer here the codes should scan nr times and change the
> scanning amount from nr-1
> to nr?

Let me make it clearer. if nr=5, the original code will loop 5 times,
but in the 5th loop, it returns directly, so __select_idle_cpu is
only done 4 times.

if nr=1, the original code will loop 1 time, but in the 1st loop,
it returns directly, so __select_idle_cpu is done 0 times.

if i change the code to if(!nr--), while nr=5, the new code will
do __select_idle_cpu() 5 times rather than 4 times in the
original code.

but of course the new code changes the __select_idle_cpu
from zero to one time for the corner case nr==1.
>
> Thanks
> Barry

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