sched: Simplify the free path some
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@chello.nl>
Cc: Mike Galbraith <efault@gmx.de>
Cc: Nick Piggin <npiggin@kernel.dk>
Cc: Linus Torvalds <torvalds@linux-foundation.org>
Cc: Andrew Morton <akpm@linux-foundation.org>
Link: http://lkml.kernel.org/r/20110407122942.298339503@chello.nl
Signed-off-by: Ingo Molnar <mingo@elte.hu>
diff --git a/kernel/sched.c b/kernel/sched.c
index 6520484..72c194c 100644
--- a/kernel/sched.c
+++ b/kernel/sched.c
@@ -7023,7 +7023,8 @@
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:
@@ -7208,7 +7209,7 @@
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)
@@ -7261,12 +7262,10 @@
}
rcu_read_unlock();
- __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 cpumask_var_t *doms_cur; /* current sched domains */