[PATCH 2/2] sched/topology: optimize topology_span_sane()
From: Yury Norov
Date: Wed Aug 07 2024 - 15:05:59 EST
The function may call cpumask_equal with tl->mask(cpu) == tl->mask(i),
even though cpu != i. In such case, cpumask_equal() would always return
true, and we can proceed to the next iteration immediately.
Comment is provided by Valentin Schneider.
Reviewed-by: Valentin Schneider <vschneid@xxxxxxxxxx>
Signed-off-by: Yury Norov <yury.norov@xxxxxxxxx>
---
kernel/sched/topology.c | 12 ++++++++++++
1 file changed, 12 insertions(+)
diff --git a/kernel/sched/topology.c b/kernel/sched/topology.c
index 8af3b48da458..3661d4173d1f 100644
--- a/kernel/sched/topology.c
+++ b/kernel/sched/topology.c
@@ -2370,6 +2370,18 @@ static bool topology_span_sane(struct sched_domain_topology_level *tl,
*/
for_each_cpu_from(i, cpu_map) {
mi = tl->mask(i);
+ /*
+ * Some topology levels (e.g. PKG in default_topology[])
+ * have a sched_domain_mask_f implementation that reuses
+ * the same mask for several CPUs (in PKG's case, one mask
+ * for all CPUs in the same NUMA node).
+ *
+ * For such topology levels, repeating cpumask_equal()
+ * checks is wasteful. Instead, we first check that the
+ * tl->mask(i) pointers aren't the same.
+ */
+ if (mi == mc)
+ continue;
/*
* We should 'and' all those masks with 'cpu_map' to exactly
--
2.43.0