Re: Why Warning?

["Alexei Pytel" <pytel@xxxxxxxxxx>]

If you want fully prototyped function without parameters in C, you should
code
     void f( void );

    Alexei Pytel



"Leif Svalgaard" <leif@leif.org>@midrange.com on 01/17/2001 06:20:34 PM

Please respond to MIDRANGE-L@midrange.com

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To:   <MIDRANGE-L@midrange.com>
cc:
Subject:  Re: Why Warning?



From: Alexei Pytel <pytel@us.ibm.com>
> By definition, in C, specifying void f() means a function which does not
> return any value, but can take any number of parameters of any type. So
the
> compiler is right - this is not complete specification.

Then the message should have been" "Incomplete Specification" and
not "Not Prototyped".
Furthermore the intent was just that: no return value, and any number
of parameters including the actual number, zero. What is incomplete
about that?

>
> By contrast, in C++ void f() means function with NO parameters, which is
> complete specification.
>
>     Alexei Pytel
>
>
>
> "Leif Svalgaard" <leif@leif.org>@midrange.com on 01/17/2001 06:39:53 AM
>
> Please respond to MIDRANGE-L@midrange.com
>
> Sent by:  owner-midrange-l@midrange.com
>
>
> To:   <MIDRANGE-L@midrange.com>
> cc:
> Subject:  Re: Why Warning?
>
>
>
> From: Leif Svalgaard <leif@leif.org>
>
> > Folks, the following program works, but why do I get the warning?
> >     1       |#include <stdio.h>
> >     2       |void f();
> >     3       |
> >     4       |main() {
> >     5       |  f();
> > ===========>...a...
> > *=Warning=========> a - CZM0185 Function f has not been prototyped
prior
> to
> use.
> >     6       |}
> >     7       |
> >     8       |void f() {
> >     9       |  printf("hello\n");
> >   10       |}
>
>
> ====> use f(void) instead of f(), but it is crummy warning text.
>
>
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