Re: Are pointers to be supposed to be sign or zero extended to wider integers?

[Richard Guenther <richard dot guenther at gmail dot com>]

On Fri, Feb 12, 2010 at 10:41 AM, Jakub Jelinek <jakub@redhat.com> wrote:
> Hi!
>
> It seems pointers are sign extended to wider integers, is that intentional?
> It certainly contradicts the comment in convert_to_integer:
> ?switch (TREE_CODE (intype))
> ? ?{
> ? ?case POINTER_TYPE:
> ? ?case REFERENCE_TYPE:
> ? ? ?if (integer_zerop (expr))
> ? ? ? ?return build_int_cst (type, 0);
>
> ? ? ?/* Convert to an unsigned integer of the correct width first, and from
> ? ? ? ? there widen/truncate to the required type. ?Some targets support the
> ? ? ? ? coexistence of multiple valid pointer sizes, so fetch the one we need
> ? ? ? ? from the type. ?*/
> ? ? ?expr = fold_build1 (CONVERT_EXPR,
> ? ? ? ? ? ? ? ? ? ? ? ? ?lang_hooks.types.type_for_size
> ? ? ? ? ? ? ? ? ? ? ? ? ? ?(TYPE_PRECISION (intype), 0),
> ? ? ? ? ? ? ? ? ? ? ? ? ?expr);
> ? ? ?return fold_convert (type, expr);
> but the comment is newer than the sign extension.
>
> void
> foo (long long l)
> {
> ?if ((l >> (sizeof (void *) * __CHAR_BIT__ - 1)) == 1)
> ? ?__builtin_puts ("pointers zero extend to wider integers");
> ?else if ((l >> (sizeof (void *) * __CHAR_BIT__ - 1)) == -1)
> ? ?__builtin_puts ("pointers sign extend to wider integers");
> }
>
> int
> main (void)
> {
> ?int i;
> ?if (sizeof (&i) < sizeof (long long))
> ? ?foo ((long long) &i);
> ?return 0;
> }

Your program prints zero-extends for ICC.

Probably the behavior is undefined and we get a warning anyway:

t.c: In function ‘main’:
t.c:15: warning: cast from pointer to integer of different size

The middle-end requires an intermediate conversion to a same-precision
integer type to not fall into the trap deciding what sign a
pointer has.

Richard.

> ? ? ? ?Jakub
>