On Wed 17-10-12 16:28:38, David Rientjes wrote:

On Tue, 16 Oct 2012, Glauber Costa wrote:

[...]

quoted

+
+static void memcg_kmem_mark_dead(struct mem_cgroup *memcg)
+{
+	if (test_bit(KMEM_ACCOUNTED_ACTIVE, &memcg->kmem_accounted))
+		set_bit(KMEM_ACCOUNTED_DEAD, &memcg->kmem_accounted);
+}

The set_bit() doesn't happen atomically with the test_bit(), what 
synchronization is required for this?

The group has to be active in order to become dead so the ordering is
natural and you do not need to test&set atomicaly. Also once a group
becomes active it is always marked that way until it goes away.
-- 
Michal Hocko
SUSE Labs