Re: Plain accesses and data races in the Linux Kernel Memory Model

Paul E. McKenney
2019-01-18T18:53:59Z

On Thu, Jan 17, 2019 at 02:43:54PM -0500, Alan Stern wrote:
> On Wed, 16 Jan 2019, Andrea Parri wrote:
>
> > Can the compiler (maybe, it does?) transform, at the C or at the "asm"
> > level, LB1's P0 in LB2's P0 (LB1 and LB2 are reported below)?
> >
> > C LB1
> >
> > {
> > int *x = &a;
> > }
> >
> > P0(int **x, int *y)
> > {
> > int *r0;
> >
> > r0 = rcu_dereference(*x);
> > *r0 = 0;
> > smp_wmb();
> > WRITE_ONCE(*y, 1);
> > }
> >
> > P1(int **x, int *y, int *b)
> > {
> > int r0;
> >
> > r0 = READ_ONCE(*y);
> > rcu_assign_pointer(*x, b);
> > }
> >
> > exists (0:r0=b /\ 1:r0=1)
> >
> >
> > C LB2
> >
> > {
> > int *x = &a;
> > }
> >
> > P0(int **x, int *y)
> > {
> > int *r0;
> >
> > r0 = rcu_dereference(*x);
> > if (*r0)
> > *r0 = 0;
> > smp_wmb();
> > WRITE_ONCE(*y, 1);
> > }
> >
> > P1(int **x, int *y, int *b)
> > {
> > int r0;
> >
> > r0 = READ_ONCE(*y);
> > rcu_assign_pointer(*x, b);
> > }
> >
> > exists (0:r0=b /\ 1:r0=1)
> >
> > LB1 and LB2 are data-race free, according to the patch; LB1's "exists"
> > clause is not satisfiable, while LB2's "exists" clause is satisfiable.
>
> Umm. Transforming
>
> *r0 = 0;
>
> to
>
> if (*r0 != 0)
> *r0 = 0;
>
> wouldn't work on Alpha if r0 was assigned from a plain read with no
> memory barrier between. But when r0 is assigned from an
> rcu_dereference call, or if there's no indirection (as in "if (a != 0)
> a = 0;"), the compiler is indeed allowed to perform this
> transformation.
>
> This means my definition of preserved writes was wrong; a write we
> thought had to be preserved could instead be transformed into a read.
>
> This objection throws a serious monkey wrench into my approach. For
> one thing, it implies that (as in the example) we can't expect
> smp_wmb() always to order plain writes. For another, it means we have
> to assume a lot more writes need not be preserved.
>
> I don't know. This may doom the effort to formalize dependencies to
> plain accesses. Or at least, those other than address dependencies
> from marked reads.

(Catching up, hello from Auckland!)

At this point, I am very much in favor of taking the simpler starting
point. If someone is using any sort of dependency from a plain access,
all bets are off. Similarly, if someone is using a control or data
dependency even from a marked access, the later dependent access must
be marked to guarantee ordering.

I believe that the transformation from "*r0 = 0" should be convincing. ;-)

Thanx, Paul