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

[Paul E. McKenney]

On Thu, Oct 15, 2015 at 11:11:01AM +0800, Boqun Feng wrote:
> Hi Paul,
> 
> On Thu, Oct 15, 2015 at 08:53:21AM +0800, Boqun Feng wrote:
> > On Wed, Oct 14, 2015 at 02:44:53PM -0700, Paul E. McKenney wrote:
> [snip]
> > > 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?
> > 
> > If at this point P0's write to 'a' hasn't propagated then when?
> 
> Hmm.. I played around ppcmem, and figured out what happens to
> propagation of P0's write to 'a':
> 
> At this point, or some point after store 'a' to 1 and before sync on
> P1 finish, writes to 'a' reachs a coherence point which 'a' is 2, so
> P0's write to 'a' "fails" and will not propagate.
> 
> I probably misunderstood the word "propagate", which actually means an
> already coherent write gets seen by another CPU, right?

It is quite possible for a given write to take a position in the coherence
order that guarantees that no one will see it, as is the case here.
But yes, all readers will see an order of values for a given memory
location that is consistent with the coherence order.

> So my question should be:
> 
> As lwsync can order P0's write to 'a' happens after P0's write to 'x',
> why P0's write to 'x' isn't seen by P1 after P1's write to 'a' overrides
> P0's?

There is no global clock for PPC's memory model.

> But ppcmem gave me the answer ;-) lwsync won't wait under P0's write to
> 'x' gets propagated, and if P0's write to 'a' "wins" in write coherence,
> lwsync will guarantee propagation of 'x' happens before that of 'a', but
> if P0's write to 'a' "fails", there will be no propagation of 'a' from
> P0. So that lwsync can't do anything here.

I believe that this is consistent, but the corners can get tricky.

							Thanx, Paul

> Regards,
> Boqun
> 
> > 
> > > o	Commit P0's final register-to-register move.
> > > 
> > > o	Commit P1's sync instruction.
> > > 
> > > o	There is now nothing that can happen in either processor.
> > > 	P0 is done, and P1 is waiting for its sync.  Therefore,
> > > 	propagate P1's a=2 write to the coherence point and to
> > > 	the other thread.
> > > 
> > > o	There is still nothing that can happen in either processor.
> > > 	So pick the barrier propagate, then the acknowledge sync.
> > > 
> > > o	P1 can now execute its read from x.  Because P0's write to
> > > 	x is still waiting to propagate to P1, this still reads
> > > 	x=0.  Execute and commit, and we now have both r3 registers
> > > 	equal to zero and the final value a=2.
> > > 
> > > o	Clean up by propagating the write to x everywhere, and
> > > 	propagating the lwsync.
> > > 
> > > And the "exists" clause really does trigger: 0:r3=0; 1:r3=0; [a]=2;
> > > 
> > > I am still not 100% confident of my litmus test.  It is quite possible
> > > that I lost something in translation, but that is looking less likely.
> > >