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Subject[PATCH 04/14] sched/topology: Simplify build_overlap_sched_groups()
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@infradead.org>
---
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;


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