Re: while_each_thread() under rcu_read_lock() is broken?

From: Paul E. McKenney
Date: Thu Jun 24 2010 - 23:42:44 EST

On Thu, Jun 24, 2010 at 05:08:10PM -0700, Eric W. Biederman wrote:
> Oleg Nesterov <oleg@xxxxxxxxxx> writes:
> > On 06/24, Chris Friesen wrote:
> >>
> >> On 06/24/2010 12:07 PM, Paul E. McKenney wrote:
> >>
> >> > 3. The thread-group leader might do pthread_exit(), removing itself
> >> > from the thread group -- and might do so while the hapless reader
> >> > is referencing that thread.
> >> >
> >> > But isn't this prohibited? Or is it really legal to do a
> >> > pthread_create() to create a new thread and then have the
> >> > parent thread call pthread_exit()? Not something I would
> >> > consider trying in my own code! Well, I might, just to
> >> > be perverse, but... ;-)
> >>
> >> I believe SUS allows the main thread to explicitly call pthread_exit(),
> >> leaving the other threads to run. If the main() routine just returns
> >> then it implicitly calls exit().
> >
> > Correct.
> >
> > But, to clarify, if the main thread does pthread_exit() (sys_exit,
> > actually), it won't be removed from the group. It will be zombie
> > until all other threads exit.
> That we don't cleanup that zombie leaders is unfortunate really, it
> means we have the entire de_thread special case. But short fixing
> libpthread to not make bad assumptions there is little we can do about
> it really.

Keeping the zombie leaders does make at least one of the lockless
scan cases quite a bit simpler. I think, anyway.

> I'm only half following this conversation.
> If what we are looking for is a stable list_head that won't disappear
> on us we should be able to put one in sighand_struct or signal_struct
> (I forget which is which at the moment) and have a list_head that
> lives for the life of the longest living thread, and that won't get
> messed up by things like de_thread, and then next_thread could simply
> return NULL when we hit the end of the list.

Oleg did suggest this possibility, but there were complications that
I do not claim to fully understand.

Thanx, Paul
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