Re: [PATCH v2 2/2] tools/memory-model: Make ppo a subrelation of po

["Paul E. McKenney" <paulmck@kernel.org>]

On Sat, Jan 28, 2023 at 05:59:52PM -0500, Alan Stern wrote:
> On Sat, Jan 28, 2023 at 11:14:17PM +0100, Andrea Parri wrote:
> > > Evidently the plain-coherence check rules out x=1 at the 
> > > end, because when I relax that check, x=1 becomes a possible result.  
> > > Furthermore, the graphical output confirms that this execution has a 
> > > ww-incoh edge from Wx=2 to Wx=1.  But there is no ww-vis edge from Wx=1 
> > > to Wx=2!  How can this be possible?  It seems like a bug in herd7.
> > 
> > By default, herd7 performs some edges removal when generating the
> > graphical outputs.  The option -showraw can be useful to increase
> > the "verbosity", for example,
> > 
> >   [with "exists (x=2)", output in /tmp/T.dot]
> >   $ herd7 -conf linux-kernel.cfg T.litmus -show prop -o /tmp -skipchecks plain-coherence -doshow ww-vis -showraw ww-vis
> 
> Okay, thanks, that helps a lot.
> 
> So here's what we've got.  The litmus test:
> 
> 
> C hb-and-int
> {}
> 
> P0(int *x, int *y)
> {
>     *x = 1;
>     smp_store_release(y, 1);
> }
> 
> P1(int *x, int *y, int *dx, int *dy, spinlock_t *l)
> {
>     spin_lock(l);
>     int r1 = READ_ONCE(*dy);
>     if (r1==1)
>         spin_unlock(l);
> 
>     int r0 = smp_load_acquire(y);
>     if (r0 == 1) {
>         WRITE_ONCE(*dx,1);
>     }

The lack of a spin_unlock() when r1!=1 is intentional?

It is admittedly a cute way to prevent P3 from doing anything
when r1!=1.  And P1 won't do anything if P3 runs first.

> }
> 
> P2(int *dx, int *dy)
> {
>     WRITE_ONCE(*dy,READ_ONCE(*dx));
> }
> 
> 
> P3(int *x, spinlock_t *l)
> {
>     spin_lock(l);
>     smp_mb__after_unlock_lock();
>     *x = 2;
> }
> 
> exists (x=2)
> 
> 
> The reason why Wx=1 ->ww-vis Wx=2:
> 
> 	0:Wx=1 ->po-rel 0:Wy=1 and po-rel < fence < ww-post-bounded.
> 
> 	0:Wy=1 ->rfe 1:Ry=1 ->(hb* & int) 1:Rdy=1 and
> 		(rfe ; hb* & int) <= (rfe ; xbstar & int) <= vis.
> 
> 	1:Rdy=1 ->po 1:unlock ->rfe 3:lock ->po 3:Wx=2
> 		so 1:Rdy=1 ->po-unlock-lock-po 3:Wx=2
> 		and po-unlock-lock-po <= mb <= fence <= w-pre-bounded.
> 
> Finally, w-post-bounded ; vis ; w-pre-bounded <= ww-vis.
> 
> This explains why the memory model says there isn't a data race.  This 
> doesn't use the smp_mb__after_unlock_lock at all.

You lost me on this one.

Suppose that P3 starts first, then P0.  P1 is then stuck at the
spin_lock() because P3 does not release that lock.  P2 goes out for a
pizza.

Why can't the two stores to x by P0 and P3 conflict, resulting in a
data race?

							Thanx, Paul