Re: [PATCH] warnkill trivia 2/2

[Mikael Pettersson <mikpe@csd.uu.se>]

On Sun, 1 Sep 2002 14:39:03 +0200, Tomas Szepe wrote:
>I've been playing a bit with how gcc handles the const qualifiers
>and made an interesting discovery:
>
>Trying to compile
>
>typedef int *p_int;
>void a(const p_int t) { *t = 0; }
>void b(const p_int t) { t = (int *) 0; }
>void c(const int *t) { *t = 0; }
>void d(const int *t) { t = (int *) 0; }
>void e(int const *t) { *t = 0; }
>void f(int const *t) { t = (int *) 0; }
>
>will give 'assignment of read-only location' warnings for
>b(), c() and e(),

In b() t is a const value and you're trying to assign to it,
and in c() and e() t is a pointer-to-const and you're trying
to assign to *t. The compiler catches this. What's the problem?

>i.e. it's impossible to have a constant
>pointer to a non-constant value w/o using a qualified
>typedef.

void g(int * const t) { *t = 0; }

>W/o a typedef, gcc seems unable to tell the difference
>between 'const int *' and 'int const *' altogether.

There is no difference. Read the C spec, or Harbison&Steele
which has had an explanation of 'const' since their '87 2nd Ed.

/Mikael