On Fri, Mar 02, 2007 at 08:23:32PM -0800, Suresh B wrote:
> When a logical cpu 'x' already has more than one process running, then most
> likely
> the siblings of that cpu 'x' must be busy. Otherwise the idle siblings
> would have likely(in most of the scenarios) picked up the extra load making
> the load on 'x' atmost one.
Do you have any stats on this?
> Use this logic to eliminate the siblings status check and minimize the cache
> misses encountered on a heavily loaded system.
Well it does increase the cacheline footprint a bit, but all cachelines
should be local to our L1 cache, presuming you don't have any CPUs where
threads have seperate caches.
What sort of numbers do you have?
>
> Signed-off-by: Suresh Siddha <[EMAIL PROTECTED]>
> ---
>
> diff --git a/kernel/sched.c b/kernel/sched.c
> index 0dc7572..d1ecc56 100644
> --- a/kernel/sched.c
> +++ b/kernel/sched.c
> @@ -1368,7 +1368,16 @@ static int wake_idle(int cpu, struct task_struct *p)
> struct sched_domain *sd;
> int i;
>
> - if (idle_cpu(cpu))
> + /*
> + * If it is idle, then it is the best cpu to run this task.
> + *
> + * This cpu is also the best, if it has more than one task already.
> + * Siblings must be also busy(in most cases) as they didn't already
> + * pickup the extra load from this cpu and hence we need not check
> + * sibling runqueue info. This will avoid the checks and cache miss
> + * penalities associated with that.
> + */
> + if (idle_cpu(cpu) || cpu_rq(cpu)->nr_running > 1)
> return cpu;
>
> for_each_domain(cpu, sd) {
-
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