On Tue, Jan 26, 2016 at 11:58:20AM -0800, Paul E. McKenney wrote:

On Tue, Jan 26, 2016 at 12:16:09PM +0000, Will Deacon wrote:

quoted

On Mon, Jan 25, 2016 at 10:03:22PM -0800, Paul E. McKenney wrote:

quoted

On Mon, Jan 25, 2016 at 04:42:43PM +0000, Will Deacon wrote:

quoted

On Fri, Jan 15, 2016 at 01:58:53PM -0800, Paul E. McKenney wrote:

quoted

PPC Overlapping Group-B sets version 4
""
(* When the Group-B sets from two different barriers involve instructions in
   the same thread, within that thread one set must contain the other.

	P0	P1	P2
	Rx=1	Wy=1	Wz=2
	dep.	lwsync	lwsync
	Ry=0	Wz=1	Wx=1
	Rz=1

	assert(!(z=2))

   Forbidden by ppcmem, allowed by herd.
*)
{
0:r1=x; 0:r2=y; 0:r3=z;
1:r1=x; 1:r2=y; 1:r3=z; 1:r4=1;
2:r1=x; 2:r2=y; 2:r3=z; 2:r4=1; 2:r5=2;
}
 P0		| P1		| P2		;
 lwz r6,0(r1)	| stw r4,0(r2)	| stw r5,0(r3)	;
 xor r7,r6,r6	| lwsync	| lwsync	;
 lwzx r7,r7,r2	| stw r4,0(r3)	| stw r4,0(r1)	;
 lwz r8,0(r3)	|		|		;

exists
(z=2 /\ 0:r6=1 /\ 0:r7=0 /\ 0:r8=1)

That really hurts. Assuming that the "assert(!(z=2))" is actually there
to constrain the coherence order of z to be {0->1->2}, then I think that
this test is forbidden on arm using dmb instead of lwsync. That said, I
also don't think the Rz=1 in P0 changes anything.

What about the smp_wmb() variant of dmb that orders only stores?

Tricky, but I think it still works out if the coherence order of z is as
I described above. The line of reasoning is weird though -- I ended up
considering the two cases where P0 reads z before and after it reads x
and what that means for the read of y.

By "works out" you mean that ARM prohibits the outcome?

Yes, that's my understanding.

Will