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@redhat.com>
Signed-off-by: Yury Norov <yury.norov@gmail.com>
---
 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