Re: [PATCH 01/10] sched: Remove one division operation in find_busiest_queue()

[Peter Zijlstra <peterz@infradead.org>]

On Thu, Aug 22, 2013 at 01:58:28AM -0700, Paul Turner wrote:

> >         wl_i / power_i > wl_j / power_j :=
> >         wl_i * power_j > wl_j * power_i
> >
> >         struct rq *busiest = NULL, *rq;
> > -       unsigned long max_load = 0;
> > +       unsigned long busiest_load = 0, busiest_power = SCHED_POWER_SCALE;
> 
> Initializing this to SCHED_POWER_SCALE assigns a meaning that isn't
> really there.  How about just 1?

Right, 1 works, all we really need is for wl to be > 0.

> >         int i;
> >
> >         for_each_cpu(i, sched_group_cpus(group)) {
> > @@ -5049,10 +5049,9 @@ static struct rq *find_busiest_queue(str
> >                  * the load can be moved away from the cpu that is potentially
> >                  * running at a lower capacity.
> >                  */
> > -               wl = (wl * SCHED_POWER_SCALE) / power;
> > -
> > -               if (wl > max_load) {
> > -                       max_load = wl;
> 
> A comment wouldn't hurt here.

Agreed, something like so?

/*
 * Since we're looking for max(wl_i / power_i) crosswise multiplication
 * to rid ourselves of the division works out to:
 * wl_i * power_j > wl_j * power_i;  where j is our previous maximum.
 */

> > +               if (wl * busiest_power > busiest_load * power) {
> > +                       busiest_load = wl;
> > +                       busiest_power = power;
> >                         busiest = rq;
> >                 }
> >         }
> >
> >