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From: fnf@mcdsun.UUCP (Fred Fish)
Newsgroups: comp.sys.amiga
Subject: Re: Manx C
Message-ID: <351@mcdsun.UUCP>
Date: Thu, 6-Aug-87 12:01:21 EDT
Article-I.D.: mcdsun.351
Posted: Thu Aug  6 12:01:21 1987
Date-Received: Sat, 8-Aug-87 12:09:56 EDT
References: <4540@jade.BERKELEY.EDU> <1836@vax135.UUCP> <4595@jade.BERKELEY.EDU> <3199@zen.berkeley.edu>
Reply-To: fnf@mcdsun.UUCP (Fred Fish)
Distribution: comp
Organization: Motorola Microcomputer Division
Lines: 38
Keywords: look at the assembly.

In article <3199@zen.berkeley.edu> waterman@cory.Berkeley.EDU.UUCP (T.S. Alan Waterman) writes:
>Mike, the deal with this problem is this:
>
>  if (ptr == 0)  compares the pointer with an INT, which is gonna be 16 bits.
>    That is, is hacks ptr in half and compares the bottom half with 0 (16 bits).

No, Mike was right.  Bear in mind that I haven't actually tried the example
given to see what the generated code is, but if it is as you described
above then it is broken.

From draft proposed Ansi C standard, 3.2.2.3:

	"An integral constant expression with the value 0, or such an 
	expression cast to type void *, is called a null pointer constant.
	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.  ..."

In other words, the 16 bit int is widened to 32 bits, not the other
way around.  Now granted, Manx C is hardly "Ansi C", but the above
rule is not something new with Ansi C.

>
>In short, the error message is correct. You are being rightly warned the the
>compiler is about to do exactly what you told it to, and not really what
>you wanted to do. Oh, BTW, defining anything as a ((void *) x) makes a long-
>word (32 bit) value, whether x is long or not.

If it behaves as you believe, the compiler is rightly warning that it
it doing whatever it feels like, not what you told it to, or what it should
be doing.  :-)

-Fred

-- 
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= Fred Fish  Motorola Computer Division, 3013 S 52nd St, Tempe, Az 85282  USA
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