On Wed, Jul 11, 2018 at 09:06:49AM -0400, Steven Rostedt wrote:
> On Wed, 11 Jul 2018 14:56:47 +0200
> Peter Zijlstra <peterz@xxxxxxxxxxxxx> 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.
thanks!
- Joel
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