[PATCH v3 2/3] sched/topology: optimize topology_span_sane()

From: Yury Norov
Date: Mon Sep 02 2024 - 14:36:45 EST


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@xxxxxxxxxx>
Signed-off-by: Yury Norov <yury.norov@xxxxxxxxx>
---
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