Path: utzoo!utgpu!jarvis.csri.toronto.edu!mailrus!ames!nap1!ark1!dtix!mimsy!chris
From: chris@mimsy.UUCP (Chris Torek)
Newsgroups: comp.lang.c
Subject: Re: use of if (!cptr) and if (cptr), where cptr is a *
Message-ID: <18727@mimsy.UUCP>
Date: 25 Jul 89 05:21:52 GMT
References: <10099@mpx2.mpx.com> <93@microsoft.UUCP> <10100@mpx2.mpx.com> <1989Jul24.194646.3012@nc386.uucp>
Organization: U of Maryland, Dept. of Computer Science, Coll. Pk., MD 20742
Lines: 30

In article <1989Jul24.194646.3012@nc386.uucp> jeffl@nc386.uucp (Jeff Leyser)
writes:
>OK, this may be meaningless, but out of curiosity is:
>
>	if (cptr == (int) 0)
>
>illegal C, or simply compleat garbage?

[where cptr was declared with `char *cptr;']

If it has a meaning, it means to compare cptr against a nil of type
pointer-to-char.  The question comes down to `is (int)0 an integral
constant expression with value zero', because the way one writes the
untyped nil pointer in C is to write an integral constant expression
whose value is zero.

Clearly (int)0 is an integral expression whose values is zero.  Whether
the cast takes away its `constant-ness' is less certain.  Different
compilers have had differing opinions in the past; what the current
pANS says I am not sure (my copy is one or two revisions out of date,
and is elsewhere at the moment anyway).

The short answer is

	if (cptr == (int)0)

is probably meaningful garbage. :-)
-- 
In-Real-Life: Chris Torek, Univ of MD Comp Sci Dept (+1 301 454 7163)
Domain:	chris@mimsy.umd.edu	Path:	uunet!mimsy!chris