Re: [PATCH 1/8] cgroup: implement cgroup_is_ancestor()

Michal Hocko
2013-04-09T06:42:53Z

On Mon 08-04-13 14:36:46, Tejun Heo wrote:
> On Mon, Apr 08, 2013 at 08:03:44PM +0200, Michal Hocko wrote:
> > __mem_cgroup_same_or_subtree relies on css_is_ancestor if hierarchy is
> > enabled for ages. This, however, is not correct because use_hierarchy
> > doesn't need to be true all the way up the cgroup hierarchy. Consider
> > the following example:
> > root (use_hierarchy=0)
> > \
> > A (use_hierarchy=0)
> > \
> > B (use_hierarchy=1)
> > \
> > C (use_hierarchy=1)
> >
> > __mem_cgroup_same_or_subtree(A, C) would return true even though C is
> > not from the same hierarchy subtree. The bug shouldn't be critical but
> > at least dump_tasks might print unrelated tasks (via
> > task_in_mem_cgroup).
>
> Huh? Isn't that avoided by the !root_memcg->use_hierarchy test?

Yes, it is. My selective blindness strikes again :/ I was convinced that
it was memcg we tested use_hierarchy for...
Sorry about all the churn.

> > @@ -1470,9 +1470,12 @@ bool __mem_cgroup_same_or_subtree(const struct mem_cgroup *root_memcg,
> > {
> > if (root_memcg == memcg)
> > return true;
> > - if (!root_memcg->use_hierarchy || !memcg)
> ^^^^^^^^^^^^^^^^^^^^^^^^^^
> > + if (!memcg)
> > return false;
> > - return css_is_ancestor(&memcg->css, &root_memcg->css);
> > + while ((memcg = parent_mem_cgroup(memcg)))
> > + if (memcg == root_memcg)
> > + return true;
> > + return false;
> > }
> >
--
Michal Hocko
SUSE Labs