Re: Compiling GCC With a C++ Compiler (g++)

[Zack Weinberg <zack at codesourcery dot com>]

Gabriel Dos Reis <gdr@cs.tamu.edu> writes:

> Zack Weinberg <zack@codesourcery.com> writes:
> | Gabriel Dos Reis <gdr@cs.tamu.edu> writes:
> | > And how do you create it?
> | 
> | By casting away the const, as is done in build_string.  Or by
> | constructing the data in memory before giving it the type with the
> | const qualifier.
>
> Constructing the data in memory before giving it the type with the const
> qualifier, is morally what C++ constructors do.

Yes.  However, a C++ constructor cannot be used here because that's
not part of the intersection of C and C++.  A way needs to be found to
declare and initialize the structure without losing the const and
without leaving the intersection of C and C++.

If the declaration were not ill-formed in C++, then we could find a
variant of the build_string code

>   memset (s, 0, sizeof (struct tree_common));
>   TREE_SET_CODE (s, STRING_CST);
>   TREE_STRING_LENGTH (s) = len;
>   memcpy ((char *) TREE_STRING_POINTER (s), str, len);
>   ((char *) TREE_STRING_POINTER (s))[len] = '\0';

that wasn't undefined behavior, and make everybody happy.  For
instance, I think this is well-defined in both C and C++
[GATHER_STATISTICS logic left out for clarity]

tree
build_string (int len, const char *str)
{
  char *sc = ggc_alloc (sizeof (struct tree_string) + len);
  tree s = (tree) sc;

  memset (s, 0, sizeof (struct tree_common));
  TREE_SET_CODE (s, STRING_CST);
  TREE_STRING_LENGTH (s) = len;

  sc += offsetof (struct tree_string, str);
  memcpy (sc, str, len);
  sc[len] = '\0';

  return s;
}

However, we need a way to write the structure declaration too.  You
would know better than me how that might be accomplished.  (Remember,
only the intersection of C and C++ can be used.)

zw