Re: [PATCH 1/4] rcu/nocb: Protect lazy shrinker against concurrent (de-)offloading

[Frederic Weisbecker <frederic@kernel.org>]

Le Wed, Mar 22, 2023 at 04:18:24PM -0700, Paul E. McKenney a écrit :
> > @@ -1336,13 +1336,25 @@ lazy_rcu_shrink_scan(struct shrinker *shrink, struct shrink_control *sc)
> >  	unsigned long flags;
> >  	unsigned long count = 0;
> >  
> > +	/*
> > +	 * Protect against concurrent (de-)offloading. Otherwise nocb locking
> > +	 * may be ignored or imbalanced.
> > +	 */
> > +	mutex_lock(&rcu_state.barrier_mutex);
> 
> I was worried about this possibly leading to out-of-memory deadlock,
> but if I recall correctly, the (de-)offloading process never allocates
> memory, so this should be OK?

Good point. It _should_ be fine but like you, Joel and Hillf pointed out
it's asking for trouble.

We could try Joel's idea to use mutex_trylock() as a best effort, which
should be fine as it's mostly uncontended.

The alternative is to force nocb locking and check the offloading state
right after. So instead of:

	rcu_nocb_lock_irqsave(rdp, flags);
	//flush stuff
	rcu_nocb_unlock_irqrestore(rdp, flags);

Have:

	raw_spin_lock_irqsave(rdp->nocb_lock, flags);
	if (!rcu_rdp_is_offloaded(rdp))
		raw_spin_unlock_irqrestore(rdp->nocb_lock, flags);
		continue;
	}
	//flush stuff
	rcu_nocb_unlock_irqrestore(rdp, flags);

But it's not pretty and also disqualifies the last two patches as
rcu_nocb_mask can't be iterated safely anymore.

What do you think?

> >  	/* Snapshot count of all CPUs */
> >  	for_each_possible_cpu(cpu) {
> >  		struct rcu_data *rdp = per_cpu_ptr(&rcu_data, cpu);
> > -		int _count = READ_ONCE(rdp->lazy_len);
> > +		int _count;
> > +
> > +		if (!rcu_rdp_is_offloaded(rdp))
> > +			continue;
> 
> If the CPU is offloaded, isn't ->lazy_len guaranteed to be zero?
> 
> Or can it contain garbage after a de-offloading operation?

If it's deoffloaded, ->lazy_len is indeed (supposed to be) guaranteed to be zero.
Bypass is flushed and disabled atomically early on de-offloading and the
flush resets ->lazy_len.

Thanks.