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From: juul@diku.dk (Anders Juul Munch)
Newsgroups: comp.lang.c++,comp.std.c++
Subject: Re: Default type of "0"
Message-ID: <1991May7.162035.9247@odin.diku.dk>
Date: 7 May 91 16:20:35 GMT
References: <74609@brunix.UUCP>
Sender: juul@skinfaxe.diku.dk
Organization: Department of Computer Science, U of Copenhagen
Lines: 40

sdm@cs.brown.edu (Scott Meyers) writes:

>What is the type of the constant "0", independent of any context?  If there
>is such a default type, call it T.  Then are there any conversions in any
>of the following?

>    int i    = 0;  // conversion from T to int?
>    int *ip  = 0;  // conversion from T to int*?
>    double d = 0;  // conversion from T to double?

>The reason for asking is that it would be nice to avoid the following
>ambiguity:

>    void f(int x);     // parameter of a numerical type
>    void f(char *s);   // parameter of a pointer type

>    f(0);              // ambiguous call -- is 0 an int or a char*?
[stuff deleted]

Try: printf("%d,%d,%d\n", sizeof(int), sizeof(void*), sizeof(0));
on a system with different sizes for int and pointers. What you'll find is
that "0" is foremost an integer (which incidentally may be used in
place of a pointer :-).

But I wonder, if we have
	#define NULL 0
(which is ANSI C compliant, I don't know if C++ is any different), then
f(NULL) would call f(int)?! This seems highly unreasonable to me, and it
looks like yet another good reason why the implicit conversion of 0 to the
null pointer should be disallowed. 
Instead, NULL should be used whenever a null pointer is needed. NULL being
defined as
	#define NULL ((void*)0)
And
	int* ip = 0;
would then be illegal, and replaced by
	int* ip = NULL;


-- Anders Munch