Hello, Peter.
On Tue, Apr 02, 2013 at 12:29:42PM +0200, Peter Zijlstra wrote:
> On Tue, 2013-04-02 at 12:00 +0200, Peter Zijlstra wrote:
> > On Tue, 2013-04-02 at 18:50 +0900, Joonsoo Kim wrote:
> > >
> > > It seems that there is some misunderstanding about this patch.
> > > In this patch, we don't iterate all groups. Instead, we iterate on
> > > cpus of local sched_group only. So there is no penalty you mentioned.
> >
> > OK, I'll go stare at it again..
>
> Ah, I see, you're doing should_we_balance() _before_
> find_busiest_group() and instead you're doing another for_each_cpu() in
> there.
>
> I'd write the thing like:
>
> static bool should_we_balance(struct lb_env *env)
> {
> struct sched_group *sg = env->sd->groups;
> struct cpumask *sg_cpus, *sg_mask;
> int cpu, balance_cpu = -1;
>
> if (env->idle == CPU_NEWLY_IDLE)
> return true;
>
> sg_cpus = sched_group_cpus(sg);
> sg_mask = sched_group_mask(sg);
>
> for_each_cpu_and(cpu, sg_cpus, env->cpus) {
> if (!cpumask_test_cpu(cpu, sg_mask))
> continue;
>
> if (!idle_cpu(cpu))
> continue;
>
> balance_cpu = cpu;
> break;
> }
>
> if (balance_cpu == -1)
> balance_cpu = group_balance_cpu(sg);
>
> return balance_cpu == env->dst_cpu;
> }
Okay. It looks nice.
>
> I also considered doing the group_balance_cpu() first to avoid having
> to do the idle_cpu() scan, but that's a slight behavioural change
> afaict.
In my quick thought, we can avoid it through below way.
balance_cpu = group_balance_cpu(sg);
if (idle_cpu(balance_cpu))
return balance_cpu == env->dst_cpu;
else
do idle_cpus() scan loop
Is it your idea? If not, please let me know your idea.
Thanks.
>
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