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From: Malcolm Kay <malcolm.kay@internode.on.net>
Organization: At home
To: Sean Chittenden <sean@chittenden.org>
Date: Tue, 10 Jun 2003 19:28:30 +0930
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Subject: Re: gcc me harder: -Wconversion bug?
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On Tue, 10 Jun 2003 18:51, Sean Chittenden wrote:
> > It seems to me that this is doing exactly what is claimed for
> > -Wconversion. To quote from the gcc man page:
> >        -Wconversion
> >               Warn  if  a prototype causes a type conversion that
> >               is different from what would happen to the same ar-
> >               gument  in  the  absence  of a prototype.  ...
> >
> > Now in the absence of a prototype for f() the argument true would be
> > promoted from char/bool to int before being passed to the
> > function. With the prototype in scope it is not promoted. Different
> > argument widths so warning delivered.
>
> % cpp test.c
> # 1 "test7.c"
> # 1 "<built-in>"
> # 1 "<command line>"
> # 1 "test7.c"
> # 1 "test7.h" 1
> # 13 "test7.h"
> void f(char b);
> # 2 "test7.c" 2
>
> int
> main(int argc, char *argv[]) {
>   f((((char)1)));
>
>   return(0);
> }
>
> void
> f(char b) {
> }
>
> Am I missing something that says that there isn't the prototype of the
> same width?  Last time I checked my vision, f(char b) was the same as
> f(char b)...  :-/ or am I missing something?  I believe that gcc's
> promoting the char to an int or to some other non-1 byte width data
> type... but I'm not seeing how, where, or why.  -sc

According to the man page we are comaring what does happen with
the prototype in scope with what would have happened if the prototype=20
was not there.

You are aware theat the rules of C require that in the absence of a proto=
type
actual integer calling arguments of less width than int (usually char and=
=20
short) must be promoted to int before the call?

Malcolm