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

[Richard Guenther]

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
>