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From: volpe@underdog.crd.ge.com (Christopher R Volpe)
Newsgroups: comp.std.c
Subject: Re: prototypes required ?
Keywords: printf stdargs prototypes
Message-ID: <12826@crdgw1.crd.ge.com>
Date: 17 Oct 90 12:36:33 GMT
References: <4026@otis.oakhill.UUCP> <14164@smoke.BRL.MIL>
Sender: news@crdgw1.crd.ge.com
Reply-To: volpe@underdog.crd.ge.com (Christopher R Volpe)
Lines: 34

In article <14164@smoke.BRL.MIL>, gwyn@smoke.BRL.MIL (Doug Gwyn) writes:
|>In article <4026@otis.oakhill.UUCP> jeff@oakhill.UUCP (Jeff Enderwick)
writes:
|>>Is it legal for a compilation system to require prototypes when stdarg
|>>functions are used ? It valid for the compiler to require you to include
|>>the prototype:
|>>	int printf ( const char*, ... );
|>>before making the call:
|>>	 printf ( "hello %d worlds\n", 5 );
|>
|>Yes, you definitely must have a prototype in scope before calling a
|>variable-argument function in a strictly conforming program.

Two questions:
  1) Can you point me to a reference in the standard that says that
     old-style declarations are insufficient for a variable argument
     function in a strictly conforming program? Or are old style declarations
     insufficient for ANY kind of function in a strictly conforming
     program?

  2) Are you sure you answered the question he asked? Jeff asked if
     it is valid for the compiler to REQUIRE YOU TO include the
     prototype. To me, "require you to" means "refuse to generate
     code unless you...". That's a different issue. Unless a program
     violates a syntax rule or a semantic constraint, the compiler
     can at most issue a warning but must still generate code. I
     believe (please correct me if I'm wrong) that a program can
     fail to be strictly conforming (and even fail to be conforming)
     without violating any semantic constraints. Is this correct?
             
==================
Chris Volpe
G.E. Corporate R&D
volpecr@crd.ge.com