Path: utzoo!attcan!utgpu!jarvis.csri.toronto.edu!mailrus!ncar!ico!ism780c!haddock!karl
From: karl@haddock.ima.isc.com (Karl Heuer)
Newsgroups: comp.lang.c
Subject: Re: pointers to functions
Keywords: pointers, functions, portable
Message-ID: <14499@haddock.ima.isc.com>
Date: 1 Sep 89 13:54:04 GMT
References: <1679@hydra.gatech.EDU>
Reply-To: karl@haddock.ima.isc.com (Karl Heuer)
Organization: Interactive Systems, Cambridge, MA 02138-5302
Lines: 28

In article <1679@hydra.gatech.EDU> wj4@prism.gatech.EDU (JOYE,WILLIAM A) writes:
>is the following code portable and why or why not?

First, I'll answer the question you probably meant to ask: "When `f' has type
`pointer to function', is `f()' a valid way to invoke it, in addition to
`(*f)()'?"  The answer is "Yes, but this feature is new to ANSI C."  As has
already been mentioned, K&R does not guarantee this.  (But the conclusion
"even the gods can lie" is inappropriate.  It's merely the case that some, but
not all, pre-ANSI compilers allowed this feature as an extension.)

Now, in case you're interested, the answer to the question you asked is "No."

>extern void printf;

Even if you add the parens (I presume they were dropped by accident when you
posted -- I find it hard to believe that so many compilers would accept this
as written), the correct declaration is "extern int printf(char *, ...);" (or
just use "#include <stdio.h>", which defines it).  In pre-ANSI C, omit the
arglist.

>	void (*f)() = printf;

Likewise, the correct declaration here is "int (*f)(char *, ...) = printf", or
just "int (*f)() = printf" in pre-ANSI C.  (It is *not* legal to declare a
function void when you don't plan to use its result.  You must declare its
true type, and then (optionally) cast to void.)

Karl W. Z. Heuer (ima!haddock!karl or karl@haddock.isc.com), The Walking Lint