Re: [PATCH V2] lockdep: fix deadlock issue between lockdep and rcu

[Boqun Feng]

On Thu, Feb 01, 2024 at 12:56:36PM -0800, Bart Van Assche wrote:
> On 2/1/24 11:58, Boqun Feng wrote:
> > On Thu, Feb 01, 2024 at 09:22:20AM -0800, Bart Van Assche wrote:
> > > On 1/16/24 23:48, Zhiguo Niu wrote:
> > > >    	/*
> > > > -	 * If there's anything on the open list, close and start a new callback.
> > > > -	 */
> > > > -	call_rcu_zapped(delayed_free.pf + delayed_free.index);
> > > > +	* If there's anything on the open list, close and start a new callback.
> > > > +	*/
> > > > +	if (need_callback)
> > > > +		call_rcu(&delayed_free.rcu_head, free_zapped_rcu);
> > > 
> > > The comment above the if-statement refers to the call_rcu_zapped() function
> > > while call_rcu_zapped() has been changed into call_rcu(). So the comment is
> > > now incorrect.
> > > 
> > > Additionally, what guarantees that the above code won't be triggered
> > > concurrently from two different threads? As you may know calling call_rcu()
> > > twice before the callback has been started is not allowed. I think that can
> > > happen with the above code.
> > 
> > No, it's synchronized by the delayed_free.schedule. Only one thread/CPU
> > can schedule at a time. Or am I missing something subtle?
> 
> Only call_rcu_zapped() reads and modifies delayed_free.scheduled. Direct
> call_rcu() calls do neither read nor modify delayed_free.scheduled.

Have you checked the change in the patch? Now call_rcu_zapped() has been
splitted into two parts: preparing the callback and calling call_rcu(),
the preparing part checks and sets the delayed_free.scheduled under
graph_lock(), so only one CPU/thread will win and do the actual
call_rcu(). And the RCU callback free_zapped_rcu() will unset
delayed_free.scheduled, again under graph_lock().

If you think it's still possible, could you provide a case where the
race may happen?

Regards,
Boqun

> 
> Bart.
>