On Wed, 11 Jul 2018 17:31:00 -0700
Joel Fernandes <j...@joelfernandes.org> wrote:
> On Wed, Jul 11, 2018 at 09:06:49AM -0400, Steven Rostedt wrote:
> > On Wed, 11 Jul 2018 14:56:47 +0200
> > Peter Zijlstra <pet...@infradead.org> wrote:
> >
> > > On Thu, Jun 28, 2018 at 11:21:46AM -0700, Joel Fernandes wrote:
> > > > static inline void tracepoint_synchronize_unregister(void)
> > > > {
> > > > + synchronize_srcu(&tracepoint_srcu);
> > > > synchronize_sched();
> > > > }
> > >
> > > Given you below do call_rcu_sched() and then call_srcu(), isn't the
> > > above the wrong way around?
> >
> > Good catch!
> >
> > release_probes()
> > call_rcu_sched()
> > ---> rcu_free_old_probes() queued
> >
> > tracepoint_synchronize_unregister()
> > synchronize_srcu(&tracepoint_srcu);
> > < finishes right away >
> > synchronize_sched()
> > --> rcu_free_old_probes()
> > --> srcu_free_old_probes() queued
> >
> > Here tracepoint_synchronize_unregister() returned before the srcu
> > portion ran.
>
> But isn't the point of synchronize_rcu to make sure that we're no longer in
> an RCU read-side section, not that *all* queued callbacks already ran? So in
> that
> case, I think it doesn't matter which order the 2 synchronize functions are
> called in. Please let me know if if I missed something!
>
> I believe what we're trying to guarantee here is that no tracepoints using
> either flavor of RCU are active after tracepoint_synchronize_unregister
> returns.
Yes you are correct. If tracepoint_synchronize_unregister() is only to
make sure that there is no more trace events using the probes, then
this should work. I was focused on looking at it with release_probes()
too. So the patch is fine as is.
-- Steve
