Re: Uses for memory barriers

[Paul E. McKenney]

On Sat, Oct 21, 2006 at 03:47:31PM -0400, Alan Stern wrote:
> On Fri, 20 Oct 2006, Paul E. McKenney wrote:
> 
> > > > And the number before the colon is the CPU #, right?
> > > 
> > > No, it's an instance label (kind of like a line number).  As we've been 
> > > doing before.  The CPU number is still a subscript.
> > 
> > When I started actually doing formulas, the use of ":" conflicted with
> > the logic-expression operator ("for every x there exists y and z such
> > that...", which turns into upside-down-A x backwards-E y,z colon ...,
> > at least if I am remembering all this correctly).
> > 
> > So how about having the line number be a second subscript following the
> > CPU number?
> 
> Make it a second subscript following the close paren.  That way either 
> subscript can be omitted without confusion.

Or just use an unbound variable for the CPU if unspecified.

> > > > > Consider this example:
> > > > > 
> > > > > 	CPU 0			CPU 1
> > > > > 	-----			-----
> > > > > 	A = 1;			B = 2;
> > > > > 	mb();			mb();
> > > > > 	B = 1;			X = A + 1;
> > > > > 	...
> > > > > 	assert(!(B==2 && X==1));
> > > > > 
> > > > > The assertion cannot fail.  But to prove it in our formalism requires 
> > > > > writing  st_0(B=1) -> st_1(B=2).  In other words, CPU 1's store to B sees 
> > > > > (i.e., overwrites) CPU 0's store to B.
> > > 
> > > I wrote that the assertion cannot fail, but I'm not so sure any more.  
> > > Couldn't there be a system where CPU 1's statements run before any of CPU 
> > > 0's writes become visible to CPU 1, but nevertheless the caching hardware 
> > > decides that CPU 1's write to B "wins"?  More on this below...
> > 
> > The memory barriers require that if the caching hardware decides that
> > CPU 1's B=2 "wins", then X=A+1 must see CPU 0's A=1.  Now the assertion
> > could legally see B==2&&X==0, or B==1&&X={0,1,2}, but it cannot legally
> > see B==2&&X==1 because that would violate the constraints that the
> > memory barriers are supposed to be enforcing.
> 
> I won't say that you're wrong.  But how do you know for sure?  That is, 
> exactly which constraint does it violate and where is it written in stone 
> that the memory barriers are supposed to enforce this constraint?  After 
> all, this doesn't fit the usual st-wmb-st / ld-rmb-ld pattern.

I don't know for sure...  Yet.  Work in progress.

> > > But now I wonder...  This approach might require too much of memory 
> > > barriers.  If we have
> > > 
> > > 	st(A); mb(); st(B);
> > > 
> > > does the memory barrier really say anything about CPUs which never see one 
> > > or both of those stores?  Or does it say only that CPUs which see both 
> > > stores will see them in the correct order?
> > 
> > Only that CPUs that see both stores with appropriate use of memory barriers
> > will see them in the correct order.
> 
> Right -- that's what I meant.
> 
> > > The only way to be certain a CPU sees a store is if it does a load which
> > > returns the value of that store.  In the absence of such a criterion we
> > > don't know what really happened.  For example:
> > > 
> > > 	CPU 0		CPU 1			CPU 2
> > > 	-----		-----			-----
> > > 	A = 1;		while (B < 1) ;		while (B < 2) ;
> > > 	mb();		mb();			mb();
> > > 	B = 1;		B = 2;			assert(A==1);
> 
> > Yep, which it must for locking to work, assuming my approach to transitivity.
> > If we were using your approach to transitivity, we would have a different
> > memory-barrier condition that required an atomic operation as the final
> > link in a chain of changes to a given variable -- and in that case, we
> > would find that the assertion could fail.
> > 
> > > But we know that the assertion may fail on some systems!
> > 
> > To be determined -- the fact that we are threading the single variable B
> > through all three CPUs potentially makes a difference.  Working on the
> > empirical end.  I know the assert() always would succeed on POWER, and
> > empirical evidence agrees (fine-grained low-access-overhead synchronized
> > time-base registers make this easier!).  I believe that the assert would
> > also success on Alpha.
> 
> I overstated the case.  We don't know of any actual architectures where 
> the assertion could fail, but we have considered a possible architecture 
> where it might.  What about NUMA machines?

If it fails on a given system, that system would be unable to execute
some popular textbook algorithms.  So, no, I am not certain, but I am
reasonably confident.

> > >                                                          The reason being
> > > that CPU 0's stores might propagate slowly to CPU 2 but the paths from
> > > CPU 0 to CPU 1 and from CPU 1 to CPU 2 might be quick.  Hence CPU 0's
> > > write to A might not be visible to CPU 2 when the assertion runs, and
> > > CPU 0's write to B might never be visible to CPU 2.
> > 
> > Agreed -- if the assertion really is guaranteed to work, it is because
> > the cache-coherence protocol is required to explicitly make it be so.
> > One reason that I am not too pessimistic is that there are some popular
> > textbook algorithms that cannot work otherwise (e.g., the various locking
> > algorithms that don't use atomic instructions, possibly also some of the
> > NBS algorithms).
> > 
> > > This wouldn't violate the guarantee made by CPU 0's mb().  The guarantee 
> > > wouldn't apply to CPU 2, since CPU 2 never sees the "B = 1".
> > 
> > Again, this depends on the form of the memory-barrier guarantee --
> > the presence of a "==>" vs. a "=>" is quite important!  ;-)
> 
> Just so.
> 
> > My hope is that the CPUs are guaranteed to act the way that I hope they
> > do, so that the memory-barrier rule is given by:
> > 
> > st(A) -p> wmb -p> st(B) && ld(B) -p> rmb -p> ld(A) && st(B) => ld(B) ->
> > 	st(A) => ld(A)
> > 
> > If so, everything works nicely.  In your scenario, the memory-barrier
> > rule would instead be something like:
> > 
> > st(A) -p> wmb -p> st(B) && ld(B) -p> rmb -p> ld(A) && st(B) ==> ld(B) ->
> > 	st(A) => ld(A)
> > st_0(A) -p> wmb -p> st(B) && ld_1(B) -p> rmb -p> ld(A) && st_0(B) => ld_1(B) &&
> > 	not-exist i: st_0(B) => st_i(B) => ld_1(B) ->
> > 		st(A) => ld(A)
> 
> This is identical to the previous version, since by definition
> 
> 	st_i(B) ==> ld_j(B)  is equivalent to  st_i(B) => ld_j(B) &&
> 		not exist k: st_i(B) => st_k(B) => ld_j(B).

OK -- we were assuming slightly different definitions of "==>".  I as
assuming that if st==>ld1==>ld2, that it is not the case that "st==>ld2".
In this circumstance, your definition is certainly more convenient than
is mine.  In the case of MMIO, the situation might be reversed.

> > st(A) -p> wmb -p> st(B) && ld(B) -p> rmb -p> ld(A) && st(B) => at(B) ->
> > 	st(A) => ld(A)
> > 
> > In other words, either: (1) the st(B) must immediately precede the ld(B), or
> > (2) there can only be loads between the st(B) and ld(B) (yes, I would have
> > to add other terms to exclude atomics and so on...), or (3) the access to A
> > by CPU 1 must be an atomic operation.
> 
> (1) Yes; it's the same as saying that ld(B) returns the value of the 
> st(B) (no intervening stores).

Good.

> (2) doesn't make sense, since loads aren't part of the global ordering of
> accesses of B -- they are invisible.  (BTW, you don't need to assume as
> well that stores are blind; it's enough just to have loads be invisible.)  
> Each load sees an initial sequence of stores ending in the store whose
> value is returned by the load, but this doesn't mean that the load occurs
> between that store and the next one.

That is due to our difference in definition.  Perhaps the following
definition:  "A==>B" means either that B sees the value stored by A
or that B sees the same value as does A?

Some work will be required to see what is best.

> (3) The assumption should be that both accesses of B are atomic; it 
> doesn't matter whether the accesses of A are.

Check out the i386 default definition of spin_unlock() -- no atomic
operations.  So only the final access of B (the one corresponding to
spin_lock()) would need to be atomic.

> > Does this make sense?
> 
> Yes.  Maybe we should include these rules as an alternative set of "very
> weak" memory ordering assumptions?  For normal uses they shouldn't make
> any difference.

Let's see what the actual architectures really impose.  I would prefer to
shave the Mandlebrot set a bit more closely than absolutely needed than
vice versa.  ;-)

							Thanx, Paul
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