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Message-ID: <20100621174455.GA14886@redhat.com>
Date: Mon, 21 Jun 2010 19:44:55 +0200
From: Oleg Nesterov <oleg@...hat.com>
To: "Paul E. McKenney" <paulmck@...ux.vnet.ibm.com>
Cc: Andrew Morton <akpm@...ux-foundation.org>,
Don Zickus <dzickus@...hat.com>,
Frederic Weisbecker <fweisbec@...il.com>,
Ingo Molnar <mingo@...e.hu>,
Jerome Marchand <jmarchan@...hat.com>,
Mandeep Singh Baines <msb@...gle.com>,
Roland McGrath <roland@...hat.com>,
linux-kernel@...r.kernel.org, stable@...nel.org,
"Eric W. Biederman" <ebiederm@...ssion.com>
Subject: Re: while_each_thread() under rcu_read_lock() is broken?
On 06/21, Oleg Nesterov wrote:
>
> So, I am thinking about the first attempt
>
> #define while_each_thread(g, t) \
> while ((t = next_thread(t)) != g && pid_alive(g))
>
> again. But this means while_each_thread() can miss more threads
> than it currently can under the same conditions. Correct, but
> not good.
Not good, but correct ;) Probably it makes sense to fix the problem
anyway, then think about the more optimal fix.
static inline struct task_struct *
next_thread_careful(const struct task_struct *g, const struct task_struct *t)
{
t = next_thread(t);
/*
* this pairs with the implicit barrier between detach_pid()
* and list_del_rcu(g->thread_group) in __unhash_process(g).
*/
smp_rmb();
if (likely(pid_alive(g)))
return t;
else
return g;
}
#define while_each_thread(g, t) \
while ((t = next_thread_careful(t)) != g)
I think this should work. detach_pid() does unlock + lock at least
once and thus we have the barrier (this worth a comment or we
can add the explicit wmb() in __unhash_process).
Paul, Roland, do you see any problems from the correctness pov,
or a better fix for now?
Perhaps it also makes sense to keep the old variant renamed to
while_each_thread_locked(), I dunno.
Oleg.
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