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From: guy@sun.uucp (Guy Harris)
Newsgroups: net.micro.amiga,net.lang.c
Subject: Re: printf() code size using Astartup.obj and AMIGA.LIB only
Message-ID: <4907@sun.uucp>
Date: Wed, 9-Jul-86 13:52:45 EDT
Article-I.D.: sun.4907
Posted: Wed Jul  9 13:52:45 1986
Date-Received: Fri, 11-Jul-86 06:30:38 EDT
References: <8606270438.AA07486@pavepaws> <426@oscvax.UUCP> <84@unisoft.UUCP> <566@3comvax.UUCP> <1385@well.UUCP> <951@jade.BERKELEY.EDU>
Organization: Sun Microsystems, Inc.
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Xref: watmath net.micro.amiga:3911 net.lang.c:9802

> >	It never fails to amaze me just how many people do this:
> >		if (foo == NULL) { ... }
> >	or:
> >		if (foo == 0) { ... }
> [instead of...]
> >		if (!foo) { ... }
> 
> Well, I wrote it the way you think people ought to for 8 years before
> changing to coding the test explicitly. I changed because the (!foo)
> notation is less readable than the (foo == 0) notation.

Amen.  The three constructs given have exactly the same meaning in C (yes,
wait for the explanation, it's true) and any good compiler will produce
exactly the same code for them.  However, "!foo" is suggestive of testing
for the falsity of a Boolean (yes, I know C doesn't have a formal Boolean
type, but that's the proper way to think of something with values meaning
"true" and "false"), while "foo == NULL" is suggestive of testing whether a
pointer is null or not, and "foo == 0" is suggestive of testing whether an
integral value is zero or not.

> Also, if foo is a pointer type, the test (!foo) is wrong. NULL doesn't
> have to be zero.

Oh, yes it does.  A null pointer need not have a bit pattern consisting of
all zeroes.  However, the value of a null pointer is constructed in
C by converting the integral constant 0 to a pointer type.  In the construct
"foo == 0", the LHS is a pointer, so the compiler knows enough to convert
the "0" on the RHS to a pointer of the same type, hence a null pointer of
the appropriate type.  (No, I won't cite the K&R/ANSI C draft which says
this; seek and ye shall find.)

As such, NULL should be #defined as 0 - NOT "(char *)0", because that
causes the compiler to get annoyed when you compare an "int *" against NULL,
since you're comparing two different kinds of pointers.

> >	The reason the compiler is exhibiting the behavior you describe is
> >because you told it to.  You are asking the compiler to compare an unsigned
> >byte to the literal value of NULL.

Which should always be 0.  As such, in the example given, 

	if (var == NULL) ...

is identical to

	if (var == 0) ...

and should generate the same code; as the author of the response to the
above comment points out, there is NO reason to extend a 1-byte variable
"var" to 4 bytes when doing this.
-- 
	Guy Harris
	{ihnp4, decvax, seismo, decwrl, ...}!sun!guy
	guy@sun.com (or guy@sun.arpa)