Re: rcu_read_lock lost its compiler barrier

[Paul E. McKenney]

On Mon, Jun 03, 2019 at 01:26:26PM +0800, Herbert Xu wrote:
> On Sun, Jun 02, 2019 at 08:47:07PM -0700, Paul E. McKenney wrote:
> > 
> > 1.	These guarantees are of full memory barriers, -not- compiler
> > 	barriers.
> 
> What I'm saying is that wherever they are, they must come with
> compiler barriers.  I'm not aware of any synchronisation mechanism
> in the kernel that gives a memory barrier without a compiler barrier.

Yes, if a given synchronization mechanism requires that memory references
need to be ordered, both the compiler and the CPU must maintain that
ordering.

> > 2.	These rules don't say exactly where these full memory barriers
> > 	go.  SRCU is at one extreme, placing those full barriers in
> > 	srcu_read_lock() and srcu_read_unlock(), and !PREEMPT Tree RCU
> > 	at the other, placing these barriers entirely within the callback
> > 	queueing/invocation, grace-period computation, and the scheduler.
> > 	Preemptible Tree RCU is in the middle, with rcu_read_unlock()
> > 	sometimes including a full memory barrier, but other times with
> > 	the full memory barrier being confined as it is with !PREEMPT
> > 	Tree RCU.
> 
> The rules do say that the (full) memory barrier must precede any
> RCU read-side that occur after the synchronize_rcu and after the
> end of any RCU read-side that occur before the synchronize_rcu.
> 
> All I'm arguing is that wherever that full mb is, as long as it
> also carries with it a barrier() (which it must do if it's done
> using an existing kernel mb/locking primitive), then we're fine.

Fair enough, and smp_mb() does provide what is needed.

> > Interleaving and inserting full memory barriers as per the rules above:
> > 
> > 	CPU1: WRITE_ONCE(a, 1)
> > 	CPU1: synchronize_rcu	
> > 	/* Could put a full memory barrier here, but it wouldn't help. */
> 
> 	CPU1: smp_mb();
> 	CPU2: smp_mb();

What is CPU2's smp_mb() ordering?  In other words, what comment would
you put on each of the above smp_mb() calls?

> Let's put them in because I think they are critical.  smp_mb() also
> carries with it a barrier().

Again, agreed, smp_mb() implies barrier().

> > 	CPU2: rcu_read_lock();
> > 	CPU1: b = 2;	
> > 	CPU2: if (READ_ONCE(a) == 0)
> > 	CPU2:         if (b != 1)  /* Weakly ordered CPU moved this up! */
> > 	CPU2:                 b = 1;
> > 	CPU2: rcu_read_unlock
> > 
> > In fact, CPU2's load from b might be moved up to race with CPU1's store,
> > which (I believe) is why the model complains in this case.
> 
> Let's put aside my doubt over how we're even allowing a compiler
> to turn
> 
> 	b = 1
> 
> into
> 
> 	if (b != 1)
> 		b = 1
> 
> Since you seem to be assuming that (a == 0) is true in this case
> (as the assignment b = 1 is carried out), then because of the
> presence of the full memory barrier, the RCU read-side section
> must have started prior to the synchronize_rcu.  This means that
> synchronize_rcu is not allowed to return until at least the end
> of the grace period, or at least until the end of rcu_read_unlock.
> 
> So it actually should be:
> 
> 	CPU1: WRITE_ONCE(a, 1)
> 	CPU1: synchronize_rcu called
> 	/* Could put a full memory barrier here, but it wouldn't help. */
> 
> 	CPU1: smp_mb();
> 	CPU2: smp_mb();
> 
> 	CPU2: grace period starts
> 	...time passes...
> 	CPU2: rcu_read_lock();
> 	CPU2: if (READ_ONCE(a) == 0)
> 	CPU2:         if (b != 1)  /* Weakly ordered CPU moved this up! */
> 	CPU2:                 b = 1;
> 	CPU2: rcu_read_unlock
> 	...time passes...
> 	CPU2: grace period ends
> 
> 	/* This full memory barrier is also guaranteed by RCU. */
> 	CPU2: smp_mb();

But in this case, given that there are no more statements for CPU2,
what is this smp_mb() ordering?

							Thanx, Paul

> 	CPU1 synchronize_rcu returns
> 	CPU1: b = 2;	
> 
> Cheers,
> -- 
> Email: Herbert Xu <herbert@xxxxxxxxxxxxxxxxxxx>
> Home Page: http://gondor.apana.org.au/~herbert/
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>