Re: [PATCH v2] locks: Filter /proc/locks output on proc pid ns

Jeff Layton
2016-08-03T13:46:59Z

On Wed, 2016-08-03 at 10:35 +0300, Nikolay Borisov wrote:
> On busy container servers reading /proc/locks shows all the locks
> created by all clients. This can cause large latency spikes. In my
> case I observed lsof taking up to 5-10 seconds while processing around
> 50k locks. Fix this by limiting the locks shown only to those created
> in the same pidns as the one the proc was mounted in. When reading
> /proc/locks from the init_pid_ns show everything.
>
> > Signed-off-by: Nikolay Borisov <kernel@kyup.com>
> ---
>  fs/locks.c | 6 ++++++
>  1 file changed, 6 insertions(+)
>
> diff --git a/fs/locks.c b/fs/locks.c
> index ee1b15f6fc13..751673d7f7fc 100644
> --- a/fs/locks.c
> +++ b/fs/locks.c
> @@ -2648,9 +2648,15 @@ static int locks_show(struct seq_file *f, void *v)
>  {
> >   struct locks_iterator *iter = f->private;
> >   struct file_lock *fl, *bfl;
> > + struct pid_namespace *proc_pidns = file_inode(f->file)->i_sb->s_fs_info;
> > + struct pid_namespace *current_pidns = task_active_pid_ns(current);
>
> >   fl = hlist_entry(v, struct file_lock, fl_link);
>
> > > + if ((current_pidns != &init_pid_ns) && fl->fl_nspid

Ok, so when you read from a process that's in the init_pid_ns
namespace, then you'll get the whole pile of locks, even when reading
this from a filesystem that was mounted in a different pid_ns?

That seems odd to me if so. Any reason not to just uniformly use the
proc_pidns here?

> > > +     && (proc_pidns != ns_of_pid(fl->fl_nspid)))
> > + return 0;
> +
> >   lock_get_status(f, fl, iter->li_pos, "");
>
> >   list_for_each_entry(bfl, &fl->fl_block, fl_block)

--
Jeff Layton <jlayton@poochiereds.net>