memcg: memcg_kmem_create_cache: make memcg_name_buf statically allocated
It isn't worth complicating the code by allocating it on the first access,
because it only takes 256 bytes.
Signed-off-by: Vladimir Davydov <vdavydov@parallels.com>
Cc: Michal Hocko <mhocko@suse.cz>
Cc: Johannes Weiner <hannes@cmpxchg.org>
Signed-off-by: Andrew Morton <akpm@linux-foundation.org>
Signed-off-by: Linus Torvalds <torvalds@linux-foundation.org>
diff --git a/mm/memcontrol.c b/mm/memcontrol.c
index 7df7f59..5e2bfcc 100644
--- a/mm/memcontrol.c
+++ b/mm/memcontrol.c
@@ -3135,7 +3135,8 @@
static void memcg_kmem_create_cache(struct mem_cgroup *memcg,
struct kmem_cache *root_cache)
{
- static char *memcg_name_buf; /* protected by memcg_slab_mutex */
+ static char memcg_name_buf[NAME_MAX + 1]; /* protected by
+ memcg_slab_mutex */
struct kmem_cache *cachep;
int id;
@@ -3151,12 +3152,6 @@
if (cache_from_memcg_idx(root_cache, id))
return;
- if (!memcg_name_buf) {
- memcg_name_buf = kmalloc(NAME_MAX + 1, GFP_KERNEL);
- if (!memcg_name_buf)
- return;
- }
-
cgroup_name(memcg->css.cgroup, memcg_name_buf, NAME_MAX + 1);
cachep = kmem_cache_create_memcg(memcg, root_cache, memcg_name_buf);
/*