On Tue, Sep 15, 2015 at 10:09:41AM -0700, Paul E. McKenney wrote:
> On Tue, Sep 15, 2015 at 06:30:28PM +0200, Peter Zijlstra wrote:
> > On Tue, Sep 15, 2015 at 08:34:48AM -0700, Paul E. McKenney wrote:
> > > On Tue, Sep 15, 2015 at 04:14:39PM +0200, Peter Zijlstra wrote:
> > > > On Tue, Sep 15, 2015 at 07:09:22AM -0700, Paul E. McKenney wrote:
> > > > > On Tue, Sep 15, 2015 at 02:48:00PM +0200, Peter Zijlstra wrote:
> > > > > > On Tue, Sep 15, 2015 at 05:41:42AM -0700, Paul E. McKenney wrote:
> > > > > > > > Never mind, the PPC people will implement this with lwsync and
> > > > > > > > that is
> > > > > > > > very much not transitive IIRC.
> > > > > > >
> > > > > > > I am probably lost on context, but...
> > > > > > >
> > > > > > > It turns out that lwsync is transitive in special cases. One of
> > > > > > > them
> > > > > > > is a series of release-acquire pairs, which can extend
> > > > > > > indefinitely.
> > > > > > >
> > > > > > > Does that help in this case?
> > > > > >
> > > > > > Probably not, but good to know. I still don't think we want to rely
> > > > > > on
> > > > > > ACQUIRE/RELEASE being transitive in general though.
> > > > >
> > > > > OK, I will bite... Why not?
> > > >
> > > > It would mean us reviewing all archs (again) and documenting it I
> > > > suppose. Which is of course entirely possible.
> > > >
> > > > That said, I don't think the case at hand requires it, so lets postpone
> > > > this for now ;-)
> > >
> > > True enough, but in my experience smp_store_release() and
> > > smp_load_acquire() are a -lot- easier to use than other barriers,
> > > and transitivity will help promote their use. So...
> > >
> > > All the TSO architectures (x86, s390, SPARC, HPPA, ...) support transitive
> > > smp_store_release()/smp_load_acquire() via their native ordering in
> > > combination with barrier() macros. x86 with CONFIG_X86_PPRO_FENCE=y,
> > > which is not TSO, uses an mfence instruction. Power supports this via
> > > lwsync's partial cumulativity. ARM64 supports it in SMP via the new ldar
> > > and stlr instructions (in non-SMP, it uses barrier(), which suffices
> > > in that case). IA64 supports this via total ordering of all release
> > > instructions in theory and by the actual full-barrier implementation
> > > in practice (and the fact that gcc emits st.rel and ld.acq instructions
> > > for volatile stores and loads). All other architectures use smp_mb(),
> > > which is transitive.
> > >
> > > Did I miss anything?
> >
> > I think that about covers it.. the only odd duckling might be s390 which
> > is documented as TSO but recently grew smp_mb__{before,after}_atomic(),
> > which seems to confuse matters.
>
> Fair point, adding Martin and Heiko on CC for their thoughts.
>
> It looks like this applies to recent mainframes that have new atomic
> instructions, which, yes, might need something to make them work with
> fully transitive smp_load_acquire() and smp_store_release().
>
> Martin, Heiko, the question is whether or not the current s390
> smp_store_release() and smp_load_acquire() can be transitive.
> For example, if all the Xi variables below are initially zero,
> is it possible for all the r0, r1, r2, ... rN variables to
> have the value 1 at the end of the test.
Right... This time actually adding Martin and Heiko on CC...
Thanx, Paul
> CPU 0
> r0 = smp_load_acquire(&X0);
> smp_store_release(&X1, 1);
>
> CPU 1
> r1 = smp_load_acquire(&X1);
> smp_store_release(&X2, 1);
>
> CPU 2
> r2 = smp_load_acquire(&X2);
> smp_store_release(&X3, 1);
>
> ...
>
> CPU N
> rN = smp_load_acquire(&XN);
> smp_store_release(&X0, 1);
>
> If smp_store_release() and smp_load_acquire() are transitive, the
> answer would be "no".
>
> A similar litmus test involving atomics would be as follows, again
> with all Xi initially zero:
>
> CPU 0
> atomic_inc(&X0);
> smp_store_release(&X1, 1);
>
> CPU 1
> r1 = smp_load_acquire(&X1);
> smp_store_release(&X2, 1);
>
> CPU 2
> r2 = smp_load_acquire(&X2);
> smp_store_release(&X3, 1);
>
> ...
>
> CPU N
> rN = smp_load_acquire(&XN);
> r0 = atomic_read(&X0);
>
> Here, the question is whether r0 can be zero, but r1, r2, ... rN all
> being 1 at the end of the test.
>
> Thoughts?
>
> Thanx, Paul
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