From mboxrd@z Thu Jan 1 00:00:00 1970 Return-Path: Received: (majordomo@vger.kernel.org) by vger.kernel.org via listexpand id S932337Ab2AMP7l (ORCPT ); Fri, 13 Jan 2012 10:59:41 -0500 Received: from mx1.redhat.com ([209.132.183.28]:60664 "EHLO mx1.redhat.com" rhost-flags-OK-OK-OK-OK) by vger.kernel.org with ESMTP id S1758403Ab2AMP7h (ORCPT ); Fri, 13 Jan 2012 10:59:37 -0500 Date: Fri, 13 Jan 2012 16:20:10 +0100 From: Oleg Nesterov To: Mandeep Singh Baines Cc: Frederic Weisbecker , Li Zefan , Tejun Heo , LKML , Containers , Cgroups , KAMEZAWA Hiroyuki , Paul Menage , Andrew Morton , "Paul E. McKenney" Subject: Re: Q: cgroup: Questions about possible issues in cgroup locking Message-ID: <20120113152010.GA19215@redhat.com> References: <20111221192413.GF13529@google.com> <20111221200422.GJ17668@somewhere> <20111222153004.GA30522@redhat.com> <20120104193614.GF9511@google.com> <20120106152356.GA23995@redhat.com> <20120106182535.GJ9511@google.com> <20120111160730.GA24556@redhat.com> <20120112003102.GB9511@google.com> <20120112170728.GA25717@redhat.com> <20120112175725.GD9511@google.com> MIME-Version: 1.0 Content-Type: text/plain; charset=us-ascii Content-Disposition: inline In-Reply-To: <20120112175725.GD9511@google.com> User-Agent: Mutt/1.5.18 (2008-05-17) Sender: linux-kernel-owner@vger.kernel.org List-ID: X-Mailing-List: linux-kernel@vger.kernel.org On 01/12, Mandeep Singh Baines wrote: > > Oleg Nesterov (oleg@redhat.com) wrote: > > > > Still can't understand... Lets look at this trivial example again. > > > > We start from the main thread M, it is ->group_leader. There is > > another thread T in this thread group. We are doing > > > > OLD = M; > > > > t = M; > > do { > > do_smth(t); > > } > > while (t->group_leader == OLD && ((t = next_thread(t)) != M); > > > > The first iteration does do_smth(M). > > > > T calls de_thread() and, in particular, it does M->group_leader = T > > (see "leader->group_leader = tsk" in de_thread). > > > > after that t->group_leader == OLD fails. t == M, its group_leader == T. > > do_smth(T) won't be called. > > > > No? > > > > I think we can handle this by removing the assignment. So in de_thread(): > > - leader->group_leader = tsk; Ah, so that was you plan. I was confused by the 3rd argument, why it is needed? Yes, I thought about this too. Suppose we remove this assignment, then we can simply do #define while_each_thread(g, t) \ while (t->group_leader == g->group_leader && (t = next_thread(t)) != g) with the same effect. (to remind, currently I ignore the barriers/etc). But this can _only_ help if we start at the group leader! May be we should enforce this rule (for the lockless case), I dunno... In that case I'd prefer to add the new while_each_thread_rcu() helper. But! in this case we do not need to change de_thread(), we can simply do #define while_each_thread_rcu(t) \ while (({ t = next_thread(t); !thread_group_leader(t); })) The definition above was one of the possibilities I considered, but I wasn't able to convince myself this is the best option. See? Or do you think I missed something? Just in case... note that while_each_thread_rcu() doesn't use 'g' at all. May be it makes sense to keep the old "t != g &&", but this is minor. Oleg.