Re: [PATCH tip/locking/core v4 1/6] powerpc: atomic: Make *xchg and *cmpxchg a full barrier

[Boqun Feng <boqun.feng@gmail.com>]

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On Wed, Oct 14, 2015 at 08:07:05PM -0700, Paul E. McKenney wrote:
> On Thu, Oct 15, 2015 at 08:53:21AM +0800, Boqun Feng wrote:
[snip]
> > 
> > I'm afraid more than that, the above litmus also shows that
> > 
> > 	CPU 0				CPU 1
> > 	-----				-----
> > 
> > 	WRITE_ONCE(x, 1);		WRITE_ONCE(a, 2);
> > 	r3 = xchg_release(&a, 1);	smp_mb();
> > 					r3 = READ_ONCE(x);
> > 
> > 	(0:r3 == 0 && 1:r3 == 0 && a == 2) is not prohibitted
> > 
> > in the implementation of this patchset, which should be disallowed by
> > the semantics of RELEASE, right?
> 
> Not necessarily.  If you had the read first on CPU 1, and you had a
> similar problem, I would be more worried.
> 

Sometimes I think maybe we should say that a single unpaired ACQUIRE or
RELEASE doesn't have any order guarantee because of the above case.

But seems that's not a normal or even existing case, my bad ;-(

> > And even:
> > 
> > 	CPU 0				CPU 1
> > 	-----				-----
> > 
> > 	WRITE_ONCE(x, 1);		WRITE_ONCE(a, 2);
> > 	smp_store_release(&a, 1);	smp_mb();
> > 					r3 = READ_ONCE(x);
> > 
> > 	(1:r3 == 0 && a == 2) is not prohibitted
> > 
> > shows by:
> > 
> > 	PPC weird-lwsync
> > 	""
> > 	{
> > 	0:r1=1; 0:r2=x; 0:r3=3; 0:r12=a;
> > 	1:r1=2; 1:r2=x; 1:r3=3; 1:r12=a;
> > 	}
> > 	 P0                 | P1                 ;
> > 	 stw r1,0(r2)       | stw r1,0(r12)      ;
> > 	 lwsync             | sync               ;
> > 	 stw  r1,0(r12)	    | lwz r3,0(r2)       ;
> > 	exists
> > 	(a=2 /\ 1:r3=0)
> > 
> > Please find something I'm (or the tool is) missing, maybe we can't use
> > (a == 2) as a indication that STORE on CPU 1 happens after STORE on CPU
> > 0?
> 
> Again, if you were pairing the smp_store_release() with an smp_load_acquire()
> or even a READ_ONCE() followed by a barrier, I would be quite concerned.
> I am not at all worried about the above two litmus tests.
> 

Understood, thank you for think through that ;-)

> > And there is really something I find strange, see below.
> > 
> > > > 
> > > > So the scenario that would fail would be this one, right?
> > > > 
> > > > a = x = 0
> > > > 
> > > > 	CPU0				CPU1
> > > > 
> > > > 	r3 = load_locked (&a);
> > > > 					a = 2;
> > > > 					sync();
> > > > 					r3 = x;
> > > > 	x = 1;
> > > > 	lwsync();
> > > > 	if (!store_cond(&a, 1))
> > > > 		goto again
> > > > 
> > > > 
> > > > Where we hoist the load way up because lwsync allows this.
> > > 
> > > That scenario would end up with a==1 rather than a==2.
> > > 
> > > > I always thought this would fail because CPU1's store to @a would fail
> > > > the store_cond() on CPU0 and we'd do the 'again' thing, re-issuing the
> > > > load and now seeing the new value (2).
> > > 
> > > The stwcx. failure was one thing that prevented a number of other
> > > misordering cases.  The problem is that we have to let go of the notion
> > > of an implicit global clock.
> > > 
> > > To that end, the herd tool can make a diagram of what it thought
> > > happened, and I have attached it.  I used this diagram to try and force
> > > this scenario at https://www.cl.cam.ac.uk/~pes20/ppcmem/index.html#PPC,
> > > and succeeded.  Here is the sequence of events:
> > > 
> > > o	Commit P0's write.  The model offers to propagate this write
> > > 	to the coherence point and to P1, but don't do so yet.
> > > 
> > > o	Commit P1's write.  Similar offers, but don't take them up yet.
> > > 
> > > o	Commit P0's lwsync.
> > > 
> > > o	Execute P0's lwarx, which reads a=0.  Then commit it.
> > > 
> > > o	Commit P0's stwcx. as successful.  This stores a=1.
> > > 
> > > o	Commit P0's branch (not taken).
> > 
> > So at this point, P0's write to 'a' has propagated to P1, right? But
> > P0's write to 'x' hasn't, even there is a lwsync between them, right?
> > Doesn't the lwsync prevent this from happening?
> 
> No, because lwsync is quite a bit weaker than sync aside from just
> the store-load ordering.
> 

Understood, I've tried the ppcmem, much clear now ;-)

> > If at this point P0's write to 'a' hasn't propagated then when?
> 
> Later.  At the very end of the test, in this case.  ;-)
> 

Hmm.. I tried exactly this sequence in ppcmem, seems propagation of P0's
write to 'a' is never an option...

> Why not try creating a longer litmus test that requires P0's write to
> "a" to propagate to P1 before both processes complete?
> 

I will try to write one, but to be clear, you mean we still observe 

0:r3 == 0 && a == 2 && 1:r3 == 0 

at the end, right? Because I understand that if P1's write to 'a'
doesn't override P0's, P0's write to 'a' will propagate.

Regards,
Boqun

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