[RFC][PATCH 10/14] sched: Simplify the free path some

[Peter Zijlstra]

If we check the root_domain reference count we can see if its been
used or not, use this observation to simplify some of the return
paths.

Signed-off-by: Peter Zijlstra <a.p.zijlstra@xxxxxxxxx>
LKML-Reference: <new-submission>
---
 kernel/sched.c |   11 +++++------
 1 file changed, 5 insertions(+), 6 deletions(-)

Index: linux-2.6/kernel/sched.c
===================================================================
--- linux-2.6.orig/kernel/sched.c
+++ linux-2.6/kernel/sched.c
@@ -6999,7 +6999,8 @@ static void __free_domain_allocs(struct 
 
 	switch (what) {
 	case sa_rootdomain:
-		free_rootdomain(&d->rd->rcu); /* fall through */
+		if (!atomic_read(&d->rd->refcount))
+			free_rootdomain(&d->rd->rcu); /* fall through */
 	case sa_sd:
 		free_percpu(d->sd); /* fall through */
 	case sa_sd_storage:
@@ -7184,7 +7185,7 @@ static int __build_sched_domains(const s
 	enum s_alloc alloc_state = sa_none;
 	struct sched_domain *sd;
 	struct s_data d;
-	int i;
+	int i, ret = -ENOMEM;
 
 	alloc_state = __visit_domain_allocation_hell(&d, cpu_map);
 	if (alloc_state != sa_rootdomain)
@@ -7236,12 +7237,10 @@ static int __build_sched_domains(const s
 		sched_domain_debug(sd, i);
 	}
 
-	__free_domain_allocs(&d, sa_sd, cpu_map);
-	return 0;
-
+	ret = 0;
 error:
 	__free_domain_allocs(&d, alloc_state, cpu_map);
-	return -ENOMEM;
+	return ret;
 }
 
 static int build_sched_domains(const struct cpumask *cpu_map)


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