Re: [PATCH v2 2/7] socket: initial cgroup code.

[Kirill A. Shutemov]

On Wed, Sep 14, 2011 at 10:46:10PM -0300, Glauber Costa wrote:
> We aim to control the amount of kernel memory pinned at any
> time by tcp sockets. To lay the foundations for this work,
> this patch adds a pointer to the kmem_cgroup to the socket
> structure.
> 
> Signed-off-by: Glauber Costa <glommer@xxxxxxxxxxxxx>
> CC: David S. Miller <davem@xxxxxxxxxxxxx>
> CC: Hiroyouki Kamezawa <kamezawa.hiroyu@xxxxxxxxxxxxxx>
> CC: Eric W. Biederman <ebiederm@xxxxxxxxxxxx>
> ---
>  include/linux/memcontrol.h |   38 ++++++++++++++++++++++++++++++++++++++
>  include/net/sock.h         |    2 ++
>  net/core/sock.c            |    3 +++
>  3 files changed, 43 insertions(+), 0 deletions(-)
> 
> diff --git a/include/linux/memcontrol.h b/include/linux/memcontrol.h
> index 3b535db..be457ce 100644
> --- a/include/linux/memcontrol.h
> +++ b/include/linux/memcontrol.h
> @@ -395,5 +395,43 @@ mem_cgroup_print_bad_page(struct page *page)
>  }
>  #endif
>  
> +#ifdef CONFIG_INET
> +#ifdef CONFIG_CGROUP_MEM_RES_CTLR_KMEM
> +#include <net/sock.h>
> +static inline void sock_update_memcg(struct sock *sk)
> +{
> +	/* right now a socket spends its whole life in the same cgroup */
> +	BUG_ON(sk->sk_cgrp);
> +
> +	rcu_read_lock();
> +	sk->sk_cgrp = mem_cgroup_from_task(current);
> +
> +	/*
> +	 * We don't need to protect against anything task-related, because
> +	 * we are basically stuck with the sock pointer that won't change,
> +	 * even if the task that originated the socket changes cgroups.
> +	 *
> +	 * What we do have to guarantee, is that the chain leading us to
> +	 * the top level won't change under our noses. Incrementing the
> +	 * reference count via cgroup_exclude_rmdir guarantees that.
> +	 */
> +	cgroup_exclude_rmdir(mem_cgroup_css(sk->sk_cgrp));
> +	rcu_read_unlock();
> +}
> +
> +static inline void sock_release_memcg(struct sock *sk)
> +{
> +	cgroup_release_and_wakeup_rmdir(mem_cgroup_css(sk->sk_cgrp));
> +}

Do we really need to have these functions in the header?

-- 
 Kirill A. Shutemov
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