[v3,11/41] mips: reuse asm-generic/barrier.h

[Paul E. McKenney]

On Wed, Jan 27, 2016 at 10:25:46AM +0000, Will Deacon wrote:
> 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:
> > > On Mon, Jan 25, 2016 at 10:03:22PM -0800, Paul E. McKenney wrote:
> > > > On Mon, Jan 25, 2016 at 04:42:43PM +0000, Will Deacon wrote:
> > > > > On Fri, Jan 15, 2016 at 01:58:53PM -0800, Paul E. McKenney wrote:
> > > > > > 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.

Very good, we have agreement between the two architectures, then.  ;-)

							Thanx, Paul