Re: [PATCH v8 3/9] socket: initial cgroup code.

On Mon,  5 Dec 2011 19:34:57 -0200
Glauber Costa [off-list ref] wrote:

The goal of this work is to move the memory pressure tcp
controls to a cgroup, instead of just relying on global
conditions.

To avoid excessive overhead in the network fast paths,
the code that accounts allocated memory to a cgroup is
hidden inside a static_branch(). This branch is patched out
until the first non-root cgroup is created. So when nobody
is using cgroups, even if it is mounted, no significant performance
penalty should be seen.

This patch handles the generic part of the code, and has nothing
tcp-specific.

Signed-off-by: Glauber Costa <redacted>
CC: Kirill A. Shutemov <redacted>
CC: KAMEZAWA Hiroyuki <redacted>
CC: David S. Miller <davem-fT/PcQaiUtIeIZ0/mPfg9Q@public.gmane.org>
CC: Eric W. Biederman <ebiederm-aS9lmoZGLiVWk0Htik3J/w@public.gmane.org>
CC: Eric Dumazet <redacted>

I already replied Reviewed-by: but...

+/* Writing them here to avoid exposing memcg's inner layout */
+#ifdef CONFIG_CGROUP_MEM_RES_CTLR_KMEM
+#ifdef CONFIG_INET
+#include <net/sock.h>
+
+static bool mem_cgroup_is_root(struct mem_cgroup *memcg);
+void sock_update_memcg(struct sock *sk)
+{
+	/* A socket spends its whole life in the same cgroup */
+	if (sk->sk_cgrp) {
+		WARN_ON(1);
+		return;
+	}
+	if (static_branch(&memcg_socket_limit_enabled)) {
+		struct mem_cgroup *memcg;
+
+		BUG_ON(!sk->sk_prot->proto_cgroup);
+
+		rcu_read_lock();
+		memcg = mem_cgroup_from_task(current);
+		if (!mem_cgroup_is_root(memcg)) {
+			mem_cgroup_get(memcg);
+			sk->sk_cgrp = sk->sk_prot->proto_cgroup(memcg);
+		}
+		rcu_read_unlock();
+	}
+}

Here, you do mem_cgroup_get() if !mem_cgroup_is_root().

+EXPORT_SYMBOL(sock_update_memcg);
+
+void sock_release_memcg(struct sock *sk)
+{
+	if (static_branch(&memcg_socket_limit_enabled) && sk->sk_cgrp) {
+		struct mem_cgroup *memcg;
+		WARN_ON(!sk->sk_cgrp->memcg);
+		memcg = sk->sk_cgrp->memcg;
+		mem_cgroup_put(memcg);
+	}
+}

You don't check !mem_cgroup_is_root(). Hm, root memcg will not be freed
by this ?

Thanks,
-Kame


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