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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