linux-kernel - Re: Q: cgroup: Questions about possible issues in cgroup locking

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Message-ID: <20120112175725.GD9511@google.com>
Date:	Thu, 12 Jan 2012 09:57:25 -0800
From:	Mandeep Singh Baines <msb@...omium.org>
To:	Oleg Nesterov <oleg@...hat.com>
Cc:	Mandeep Singh Baines <msb@...omium.org>,
	Frederic Weisbecker <fweisbec@...il.com>,
	Li Zefan <lizf@...fujitsu.com>, Tejun Heo <tj@...nel.org>,
	LKML <linux-kernel@...r.kernel.org>,
	Containers <containers@...ts.linux-foundation.org>,
	Cgroups <cgroups@...r.kernel.org>,
	KAMEZAWA Hiroyuki <kamezawa.hiroyu@...fujitsu.com>,
	Paul Menage <paul@...lmenage.org>,
	Andrew Morton <akpm@...ux-foundation.org>,
	"Paul E. McKenney" <paulmck@...ux.vnet.ibm.com>
Subject: Re: Q: cgroup: Questions about possible issues in cgroup locking

Hi Oleg,

Oleg Nesterov (oleg@...hat.com) wrote:
> Hi Mandeep,
> 
> On 01/11, Mandeep Singh Baines wrote:
> > > >
> > > > #define while_each_thread(g, t, o) \
> > > > 	while (t->group_leader == o && (t = next_thread(t)) != g)
> > > >
> > > > Where o should have the value of g->group_leader.
> > >
> > > I don't understand how this helps... and how this can work even
> > > ignoring the barriers.
> > >
> > > OK, we have the main thream M and the sub-thread T, we are doing
> > >
> > > 	do {
> > > 		do_something(t);
> > > 	} while_each_thread(M, t, M);
> > >
> > > why we can't miss T if it does exec?
> > >
> >
> > So for:
> >
> > struct task *M; /* assuming this is passed in to us */
> > struct task *L = M->group_leader;
> 
> L == M
> 
> > do {
> > 	do_something(T);
> > } while_each_thread(M, T, L);
> >
> > Here is my thinking.
> >
> > If some thread K does exec, you won't miss it because:
> >
> > 1) Ignoring the group_leader check, you'll visit K just by following
> >    next_thread(). That's the case today and is what you except
> >    when iterating over an rcu_list.
> > 2) (t->group_leader == o) will fail iff t is the exec thread.
> >    Since we test t->group_leader before re-assigning it (t=next_thread()),
> >    the test will fail only after visiting the exec thread. So you'll
> >    visit the exec thread and then terminate the loop.
> 
> 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;

		tsk->exit_signal = SIGCHLD;
		leader->exit_signal = -1;

		BUG_ON(leader->exit_state != EXIT_ZOMBIE);
		leader->exit_state = EXIT_DEAD;

In the current d_thread(), four statements after reassigning
leader->group_leader, we mark the old leader as EXIT_DEAD. So what if
we leave leader->group_leader = leader. Since its EXIT_DEAD a few
statements later, I don't think anything should break.

What do you think?

Regards,
Mandeep

> Oleg.
> 
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