On Fri, Aug 18, 2023 at 10:45:56AM -0700, Yosry Ahmed wrote: > On Fri, Aug 18, 2023 at 10:35 AM Johannes Weiner <hannes@xxxxxxxxxxx> wrote: > > On Fri, Aug 18, 2023 at 07:56:37AM -0700, Yosry Ahmed wrote: > > > If this happens it seems possible for this to happen: > > > > > > cpu #1 cpu#2 > > > css_put() > > > /* css_free_rwork_fn is queued */ > > > rcu_read_lock() > > > mem_cgroup_from_id() > > > mem_cgroup_id_remove() > > > /* access memcg */ > > > > I don't quite see how that'd possible. IDR uses rcu_assign_pointer() > > during deletion, which inserts the necessary barriering. My > > understanding is that this should always be safe: > > > > rcu_read_lock() (writer serialization, in this case ref count == 0) > > foo = idr_find(x) idr_remove(x) > > if (foo) kfree_rcu(foo) > > LOAD(foo->bar) > > rcu_read_unlock() > > How does a barrier inside IDR removal protect against the memcg being > freed here though? > > If css_put() is executed out-of-order before mem_cgroup_id_remove(), > the memcg can be freed even before mem_cgroup_id_remove() is called, > right? css_put() can start earlier, but it's not allowed to reorder the rcu callback that frees past the rcu_assign_pointer() in idr_remove(). This is what RCU and its access primitives guarantees. It ensures that after "unpublishing" the pointer, all concurrent RCU-protected accesses to the object have finished, and the memory can be freed.
