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SubjectRe: [PATCH 01/10] sched: Remove one division operation in find_busiest_queue()
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;
> > }
> > }
> >
> >


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