Re: Internal vs. external barriers (was: Re: Interesting LKMM litmus test)

[Alan Stern]

On Fri, Jan 20, 2023 at 01:20:37PM -0800, Paul E. McKenney wrote:
> On Fri, Jan 20, 2023 at 03:36:24PM -0500, Alan Stern wrote:
> > On Fri, Jan 20, 2023 at 11:20:32AM -0800, Paul E. McKenney wrote:
> > > On Fri, Jan 20, 2023 at 01:37:51PM -0500, Alan Stern wrote:
> > > > srcu_read_unlock() does not need a full smp_mb().
> > > 
> > > That is quite possible, and that is what we are looking into.  And testing
> > > thus far agrees with you.  But the grace-period ordering constraints
> > > are quite severe, so this requires careful checking and severe testing.
> > 
> > If you're interested, I can provide a simple argument to show that the 
> > Fundamental Law of RCU would continue to hold with only a release fence.  
> > There is an added requirement: merely that synchronize_srcu() must have 
> > an smp_mb() somewhere after its final read of the unlock counters -- 
> > which your version of the algorithm already has.
> 
> Please!
> 
> For your amusement, here is a very informal argument that this is
> the case:
> 
> https://docs.google.com/document/d/1xvwQzavmH474MBPAIBqVyvCrCcS5j2BpqhErPhRj7Is/edit?usp=sharing
> 
> See the "Read-Side Optimizations" section at the end.

It looks like you've got the basic idea.  Most of the complications seem 
to arise from the different ways a grace period can happen.

Here's what I was thinking.  Let C be a read-side critical section, with 
L being its invocation of srcu_down_read() and U being the matching 
invocation of srcu_up_read().  Let idx be the index value read by L (and 
used by U).  I will assume that L has the form:

	idx = READ_ONCE(ss->index);
	temp = this_cpu(ss->lock)[idx];
	WRITE_ONCE(this_cpu(ss->lock)[idx], temp + 1)
	smp_mb();

(or whatever is the right syntax for incrementing a per-cpu array 
element).  Likewise, assume U has the form:

	temp = this_cpu(ss->unlock)[idx];
	smp_store_release(&this_cpu(ss->unlock)[idx], temp + 1);

Let G be any SRCU grace period -- an invocation of synchronize_srcu(ss). 
Assume G has the overall form:

	accumulate_and_compare_loop(!ss->index);
	smp_mb();
	WRITE_ONCE(ss->index, !ss->index);
	smp_mb();
	accumulate_and_compare_loop(!ss->index);

where accumulate_and_compare_loop(i) has the form:

	do {
		s = t = 0;
		for each CPU c:
			s += READ_ONCE(cpu(c, ss->unlock)[i]);
		smp_mb();
		for each CPU c:
			t += READ_ONCE(cpu(c, ss->lock)[i]);
	} while (s != t);

It's not too hard to show, and I trust you already believe, that in the 
final iteration of the accumulate_and_compare_loop(i) call for which 
i = idx, the lock-counter increment in L is observed if and only if the 
unlock-counter increment in U is observed.  Thus we have two cases:

Case 1: Both of the increments are observed.  Since the increment in U 
is a store-release, every write that propagated to U's CPU before the 
increment is therefore visible to G's CPU before its last read of an 
unlock counter.  Since the full fence in accumulate_and_compare_loop() 
is executed after the last such read, these writes must propagate to 
every CPU before G ends.

Case 2: Neither of the increments is observed.  Let W be any write which 
propagated to G's CPU before G started.  Does W propagate to C before L 
ends?  We have the following SB or RWC pattern:

	G                          C
	------------------------   -----------------------
	W propagates to G's CPU    L writes lock counter
	G does smp_mb()            L does smp_mb()
	G reads L's lock counter   W propagates to C's CPU

(The smp_mb() in the left column is the one in 
accumulate_and_compare_loop(idx), which precedes the reads of the lock 
counters.)

If L's smp_mb() ended before G's did then L's write to the lock counter 
would have propagated to G's CPU before G's smp_mb() ended, and hence G 
would have observed the lock-counter increment.  Since this didn't 
happen, we know that G's smp_mb() ends before L's does.  This means that 
W must propagate to every CPU before L terminates, and hence before C's 
critical section starts.

Together, these two cases cover the requirements of the Fundamental Law 
of RCU.  The memory barrier in U was needed only in Case 1, and there it 
only needed to be a release fence.

Alan