The function may call cpumask_equal with mi == mc, even though the CPUs are
different. In such case, cpumask_equal() would always return true, and we
can proceed to the next iteration immediately.

This happens when topologies re-use the same mask for many CPUs.
The detailed 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 | 13 +++++++++++++
 1 file changed, 13 insertions(+)

diff --git a/kernel/sched/topology.c b/kernel/sched/topology.c
index ffbe3a28d2d4..04a3b3d7b6f4 100644
--- a/kernel/sched/topology.c
+++ b/kernel/sched/topology.c
@@ -2370,6 +2370,19 @@ static bool topology_span_sane(struct sched_domain_topology_level *tl,
 	for_each_cpu_from(cpu, cpu_map) {
 		mi = tl->mask(cpu);
 
+		/*
+		 * 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
 		 * match the topology we're about to build, but that can only
-- 
2.43.0