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Re: C provenance semantics proposal


On Wed, Apr 17, 2019 at 1:53 PM Uecker, Martin
<Martin.Uecker@med.uni-goettingen.de> wrote:
>
>
> Hi Richard,
>
> Am Mittwoch, den 17.04.2019, 11:41 +0200 schrieb Richard Biener:
> > On Wed, Apr 17, 2019 at 11:15 AM Peter Sewell <Peter.Sewell@cl.cam.ac.uk> wrote:
> > >
> > > On 17/04/2019, Richard Biener <richard.guenther@gmail.com> wrote:
> > > > On Fri, Apr 12, 2019 at 5:31 PM Peter Sewell <Peter.Sewell@cl.cam.ac.uk>
> > > > wrote:
>
> ...
> > > > So this is not what GCC implements which tracks provenance through
> > > > non-pointer types to a limited extent when only copying is taking place.
> > > >
> > > > Your proposal makes
> > > >
> > > >  int a, b;
> > > >  int *p = &a;
> > > >  int *q = &b;
> > > >  uintptr_t pi = (uintptr_t)p; //expose
> > > >  uintptr_t qi = (uintptr_t)q; //expose
> > > >  pi += 4;
> > > >  if (pi == qi)
> > > >    *(int *)pi = 1;
> > > >
> > > > well-defined since (int *)pi now has the provenance of &b.
> > >
> > > Yes.  (Just to be clear: it's not that we think the above example is
> > > desirable in itself, but it's well-defined as a consequence of what
> > > we do to make other common idioms, eg pointer bit manipulation,
> > > well-defined.)
> > >
> > > > Note GCC, when tracking provenance of non-pointer type
> > > > adds like in
> > > >
> > > >   int *p = &a;
> > > >   uintptr_t pi = (uintptr_t)p;
> > > >   pi += 4;
> > > >
> > > > considers pi to have provenance "anything" (not sure if you
> > > > have something like that) since we add 4 which has provenance
> > > > "anything" to pi which has provenance &a.
> > >
> > > We don't at present have a provenance "anything", but if the gcc
> > > "anything" means that it's assumed that it might alias with anything,
> > > then it looks like gcc's implementing a sound approximation to
> > > the proposal here?
> >
> > GCC makes the code well-defined whereas the proposal would make
> > dereferencing a pointer based on pi invoke undefined behavior?
>
> No, if there is an exposed object where pi points to, it is
> defined behaviour.
>
> >  Since
> > your proposal is based on an abstract machine there isn't anything
> > like a pointer with multiple provenances (which "anything" is), just
> > pointers with no provenance (pointing outside of any object), right?
>
> This is correct. What the proposal does though is put a limit
> on where pointers obtained from integers are allowed to point
> to: They cannot point to non-exposed objects. I assume GCC
> "anything" provenances also cannot point to all possible
> objects.

Yes.  We exclude objects that do not have their address taken
though (so somewhat similar to your "exposed").

> > For points-to analysis we of course have to track all possible
> > provenances of a pointer (and if we know it doesn't point inside
> > any object we make it point to nothing).
>
> Yes, a compiler should track what it knows (it could also track
> if it knows that some pointers point to the same object, etc.)
> while the abstract machine knows everything there is to know.
>
> > Btw, GCC changed its behavior here to support optimizing matlab
> > generated C code which passes pointers to arrays across functions
> > by marshalling them in two float typed halves (yikes!).  GCC is able
> > to properly track provenance across the decomposition / recomposition
> > when doing points-to analysis ;)
>
> Impressive ;-)  I would have thought that such encoding
> happens at ABI boundaries, where you cannot track anyway.
> But this seems to occur inside compiled code?

It occurs when matlab generates C code for an expression.  They
seem to use floating-point for everything at their self-invented ABI
boundary so "obviously" that includes pointers.

> While we do not attach a provenance to integers
> in our proposal, it does not necessarily imply that a compiler
> is not allowed to track such information. It then depends on
> how it uses it.
>
> For example,
>
> int z;
> int x;
> uintptr_t pi = (uintptr_t)&x;
>
> // encode in two floats ;-)
>
> // pass floats around
>
> // decode
>
> int* p = (int*)pi;
>
> If the compiler can prove that the address is still
> the same, it can also reattach the original provenance
> under some conditions.
>
> But there is a caveat: It can only do this is it cannot
> also be  a one-after pointer for z (or some other object).
> If the address of 'z' is not exposed, it may be able to
> assume this.
>
> > Btw, one thing GCC struggles is when it applies rules that clearly
> > apply to pointer dereferences to pointer equality compares where
> > the standard has that special casing of comparing two pointers
> > where one points one after an object requiring the comparison
> > to evaluate to true when the objects are adjacent.  GCC
> > currently statically optimizes if (&x + 1 == &y) to false for
> > this reason (but not the corresponding integer comparison).
>
> Yes, according to the current rules (and this doesn't change in
> the proposal) two points comparing equal does not imply that
> they are interchangable. Making the comparison unspecified
> (as C++) would not help. We could make it undefined, which
> would make all optimizations based on the assumption that
> the pointer are interchangable valid. But I fear that this
> would introduce a corner case that could lead to subtle
> and hard-to-detect bugs.

Yeah.

Richard.

> Martin
>
> > Richard.
> >
> > >
> > > best,
> > > Peter
> > >
> > >
> > > > > The user-disambiguation refinement adds some complexity but supports
> > > > > roundtrip casts, from pointer to integer and back, of pointers that
> > > > > are one-past a storage instance.
> >
> >


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