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SubjectRe: [PATCH] sched: sched_best_cpu
On Thu, 21 Aug 2003, Frank Cornelis wrote:

> Hi,
>
>
> Next patch slightly favors the local node when looking for the least
> loaded cpu; when the task would not stay on our node we add 1 to the
> load to simulate the influence the migration would have on the load.

You might need to rediff soon due to cpumask_t changes

> --- sched.c.orig 2003-08-09 06:39:35.000000000 +0200
> +++ sched.c 2003-08-21 11:45:04.000000000 +0200
> @@ -776,29 +776,30 @@
> */
> static int sched_best_cpu(struct task_struct *p)
> {
> - int i, minload, load, best_cpu, node = 0;
> + int i, minload, load, best_cpu, node, pnode;
> unsigned long cpumask;
cpumask_t cpumask;

>
> best_cpu = task_cpu(p);
> if (cpu_rq(best_cpu)->nr_running <= 2)
> return best_cpu;
>
> - minload = 10000000;
> + minload = INT_MAX;
> + node = pnode = cpu_to_node(best_cpu);
> for_each_node_with_cpus(i) {
> /*
> * Node load is always divided by nr_cpus_node to normalise
> * load values in case cpu count differs from node to node.
> - * We first multiply node_nr_running by 10 to get a little
> - * better resolution.
> + * If the node != our node we add the load of the task.
> + * We multiply by NR_CPUS for better resolution.
> */
> - load = 10 * atomic_read(&node_nr_running[i]) / nr_cpus_node(i);
> + load = NR_CPUS * (atomic_read(&node_nr_running[i]) + (i != pnode)) / nr_cpus_node(i);
> if (load < minload) {
> minload = load;
> node = i;
> }
> }
>
> - minload = 10000000;
> + minload = INT_MAX;
> cpumask = node_to_cpumask(node);
> for (i = 0; i < NR_CPUS; ++i) {
> if (!(cpumask & (1UL << i)))

if (!cpu_isset(i, cpumask))

-
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