----- Original Message -----
From: "Josh Wilmes" <[EMAIL PROTECTED]>
To: "Brian Lee Ray" <[EMAIL PROTECTED]>
Cc: "Nicholas Clark" <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Saturday, February 23, 2002 12:08 PM
Subject: Re: [PATCH] Bowing to necessity (was Re: [PATCH]Macro bulletproofing )
> At 6:57 on 02/23/2002 CST, Brian Lee Ray <[EMAIL PROTECTED]> wrote:
> > 2:f=0;
> > f=NULL;
> > f=(void*)0; /*all valid: assignment of a null pointer
> > *constant to a function pointer*/
>
> It doesn't accept that last one.
>
> "test2.c", line 12: Error:
> [ISO 6.3.4]: Can't convert function pointer 'void *' to non-function pointer 'void
>( * ) ( void )'.
> [ISO 6.3.16]: Can't perform this conversion by assignment.
that's what I take issue with.
An integral constant expression with the value 0, or such an
expression cast to type void *, is called a null pointer
constant.
this is a null pointer constant:
(void*)0
If a null pointer constant is assigned to or compared
for equality to a pointer, the constant is converted to a
pointer of that type.
assigned to a pointer:
f=(void *)0;
compared for equality to a pointer:
f==(void*)0;
It doesn't say "to a pointer of compatible type". after all, if that were the case,
it couldn't convert the integer zero to a pointer, could it?
Just for the sake of curiosity, here's what I get with
gcc -ansi -pedantic -Wall -W -O:
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
void *v=NULL;
typedef void(*fptr)(void);
fptr f;
f=v; /*invalid: assignment of a void pointer to a function pointer*/
test2.c:10: warning: ANSI forbids assignment between function pointer and `void *'
f=0;f=NULL;f=(void*)0; /*all valid: assignment of a null pointer
*constant to a function pointer*/
v=f; /*invalid: assignment of a function pointer to a void pointer*/
test2.c:13: warning: ANSI forbids assignment between function pointer and `void *'
f==0;/*valid:comparison of a function pointer to a null pointer constant*/
test2.c:14: warning: statement with no effect
f==(void*)0;/*valid:comparison of a function pointer to a null pointer constant*/
test2.c:15: warning: statement with no effect
f==NULL;/*valid:comparison of a function pointer to a null pointer constant*/
test2.c:16: warning: statement with no effect
f==(fptr)v;/*must compile, however, unless both f and v are null pointers,
test2.c:17: warning: statement with no effect
*the result is undefined. I would expect a warning for this, as
*it is a really stupid thing to do. after all, if you know that
*f and v are both null pointers, the result will hardly be surprising*/
return 0;
}
brian.
