[PATCH 04/14] sched/topology: Simplify build_overlap_sched_groups()

[Peter Zijlstra]

Now that the first group will always be the previous domain of this
@cpu this can be simplified.

In fact, writing the code now removed should've been a big clue I was
doing it wrong :/

Signed-off-by: Peter Zijlstra (Intel) <peterz@xxxxxxxxxxxxx>
---
 kernel/sched/topology.c |   13 ++-----------
 1 file changed, 2 insertions(+), 11 deletions(-)

--- a/kernel/sched/topology.c
+++ b/kernel/sched/topology.c
@@ -557,7 +557,7 @@ static void init_overlap_sched_group(str
 static int
 build_overlap_sched_groups(struct sched_domain *sd, int cpu)
 {
-	struct sched_group *first = NULL, *last = NULL, *groups = NULL, *sg;
+	struct sched_group *first = NULL, *last = NULL, *sg;
 	const struct cpumask *span = sched_domain_span(sd);
 	struct cpumask *covered = sched_domains_tmpmask;
 	struct sd_data *sdd = sd->private;
@@ -587,15 +587,6 @@ build_overlap_sched_groups(struct sched_
 
 		init_overlap_sched_group(sd, sg, i);
 
-		/*
-		 * Make sure the first group of this domain contains the
-		 * canonical balance CPU. Otherwise the sched_domain iteration
-		 * breaks. See update_sg_lb_stats().
-		 */
-		if ((!groups && cpumask_test_cpu(cpu, sg_span)) ||
-		    group_balance_cpu(sg) == cpu)
-			groups = sg;
-
 		if (!first)
 			first = sg;
 		if (last)
@@ -603,7 +594,7 @@ build_overlap_sched_groups(struct sched_
 		last = sg;
 		last->next = first;
 	}
-	sd->groups = groups;
+	sd->groups = first;
 
 	return 0;