gcc and NULL function pointers.

[Chuck Karish]

In article <1991Jun27.190107.627 at cirrus.com> dhesi at cirrus.com
(Rahul Dhesi) writes:
>In <17605.Jun2607.39.3591 at kramden.acf.nyu.edu>
>brnstnd at kramden.acf.nyu.edu (Dan Bernstein) writes:
>>By the way, I'm curious: Why is ((char *)0) ``simply wrong''?
>
>It's wrong for the same reason that ((void *) 0) is wrong.  K&R said
>that NULL is defined to be 0.

Not in my copy.  They #defined NULL as 0 in an EXAMPLE.

>Therefore existing code that (unwisely)
>uses NULL to stand for a zero in a non-pointer context may break using
>either ((void *) 0) or ((char *) 0).

>From K&R Classic, p. 192:

    The compilers currently allow a pointer to be assigned to
    an integer, an integer to a pointer, and a pointer to a
    pointer of another type.  The assignment is a pure copy
    operation, with no conversion.  This usage is nonportable,
    and may produce pointers which cause addressing exceptions
    when used.  However, it is guaranteed that assignment odf
    the constant 0 to a pointer will produce a null pointer
    distinguishable from a pointer to any object.

I don't see how anyone could read this as prohibiting implementations
from defining NULL to be a pointer.  It does suggest that pointers
of different types may not be universally inter-assignable without
careful type casting.


>However, ((void *) 0) is blessed by ANSI and ((char *) 0) is not.
>Therefore we may summarize:
>
>     ((void *) 0) is ANSI-conformant but wrong.

It does not allow maximal portability of sloppily-written (minimalist?)
code (see K&R's warning, above), but it's not wrong.


To answer Dan's question, ((char *)0) is wrong because the
universally-castable storage type is (void *) under Standard C,
not (char *).  That's why malloc() now returns (void *).
-- 

	Chuck Karish		karish at mindcraft.com
	Mindcraft, Inc.		(415) 323-9000