Re: [PATCH v3] mm: fix race between kmem_cache destroy, create and deactivate

[Paul E. McKenney]

On Sun, Jun 10, 2018 at 07:52:50AM -0700, Shakeel Butt wrote:
> On Sat, Jun 9, 2018 at 3:20 AM Vladimir Davydov <vdavydov.dev@xxxxxxxxx> wrote:
> >
> > On Tue, May 29, 2018 at 05:12:04PM -0700, Shakeel Butt wrote:
> > > The memcg kmem cache creation and deactivation (SLUB only) is
> > > asynchronous. If a root kmem cache is destroyed whose memcg cache is in
> > > the process of creation or deactivation, the kernel may crash.
> > >
> > > Example of one such crash:
> > >       general protection fault: 0000 [#1] SMP PTI
> > >       CPU: 1 PID: 1721 Comm: kworker/14:1 Not tainted 4.17.0-smp
> > >       ...
> > >       Workqueue: memcg_kmem_cache kmemcg_deactivate_workfn
> > >       RIP: 0010:has_cpu_slab
> > >       ...
> > >       Call Trace:
> > >       ? on_each_cpu_cond
> > >       __kmem_cache_shrink
> > >       kmemcg_cache_deact_after_rcu
> > >       kmemcg_deactivate_workfn
> > >       process_one_work
> > >       worker_thread
> > >       kthread
> > >       ret_from_fork+0x35/0x40
> > >
> > > To fix this race, on root kmem cache destruction, mark the cache as
> > > dying and flush the workqueue used for memcg kmem cache creation and
> > > deactivation.
> >
> > > @@ -845,6 +862,8 @@ void kmem_cache_destroy(struct kmem_cache *s)
> > >       if (unlikely(!s))
> > >               return;
> > >
> > > +     flush_memcg_workqueue(s);
> > > +
> >
> > This should definitely help against async memcg_kmem_cache_create(),
> > but I'm afraid it doesn't eliminate the race with async destruction,
> > unfortunately, because the latter uses call_rcu_sched():
> >
> >   memcg_deactivate_kmem_caches
> >    __kmem_cache_deactivate
> >     slab_deactivate_memcg_cache_rcu_sched
> >      call_rcu_sched
> >                                             kmem_cache_destroy
> >                                              shutdown_memcg_caches
> >                                               shutdown_cache
> >       memcg_deactivate_rcufn
> >        <dereference destroyed cache>
> >
> > Can we somehow flush those pending rcu requests?
> 
> You are right and thanks for catching that. Now I am wondering if
> synchronize_sched() just before flush_workqueue() should be enough.
> Otherwise we might have to replace call_sched_rcu with
> synchronize_sched() in kmemcg_deactivate_workfn which I would not
> prefer as that would holdup the kmem_cache workqueue.
> 
> +Paul
> 
> Paul, we have a situation something similar to the following pseudo code.
> 
> CPU0:
> lock(l)
> if (!flag)
>   call_rcu_sched(callback);
> unlock(l)
> ------
> CPU1:
> lock(l)
> flag = true
> unlock(l)
> synchronize_sched()
> ------
> 
> If CPU0 has called already called call_rchu_sched(callback) then later
> if CPU1 calls synchronize_sched(). Is there any guarantee that on
> return from synchronize_sched(), the rcu callback scheduled by CPU0
> has already been executed?

No.  There is no such guarantee.

You instead want rcu_barrier_sched(), which waits for the callbacks from
all prior invocations of call_rcu_sched() to be invoked.

Please note that synchronize_sched() is -not- sufficient.  It is only
guaranteed to wait for a grace period, not necessarily for all prior
callbacks.  This goes both directions because if there are no callbacks
in the system, then rcu_barrier_sched() is within its rights to return
immediately.

So please make sure you use each of synchronize_sched() and
rcu_barrier_sched() to do the job that it was intended to do!  ;-)

If your lock(l) is shorthand for spin_lock(&l), it looks to me like you
actually only need rcu_barrier_sched():

	CPU0:
	spin_lock(&l);
	if (!flag)
	  call_rcu_sched(callback);
	spin_unlock(&l);

	CPU1:
	spin_lock(&l);
	flag = true;
	spin_unlock(&l);
	/* At this point, no more callbacks will be registered. */
	rcu_barrier_sched();
	/* At this point, all registered callbacks will have been invoked. */

On the other hand, if your "lock(l)" was instead shorthand for
rcu_read_lock_sched(), then you need -both- synchronize_sched() -and-
rcu_barrier().  And even then, you will be broken in -rt kernels.
(Which might or might not be a concern, depending on whether your code
matters to -rt kernels.

Make sense?

							Thanx, Paul