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>

mm/memcontrol.c

@@ -3135,7 +3135,8 @@ void memcg_free_cache_params(struct kmem_cache *s)
 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 @@ static void memcg_kmem_create_cache(struct mem_cgroup *memcg,
 	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);
 	/*