Re: default function parameters

[_z33 <timid.Gentoo@gmail.com>]

Steve Graegert wrote:

> Unless you're writing a compiler this does not matter.  Even if an int
> argument in implicitly used it has no meaning to the programmer. 
> Since void is a well defined type, although an incomplete one, I have
> doubts that int is used internally.  I simply can't see the rationale
> behind that (but I'd be happy to be enlightened).  Could you please
> try to transport your collegue's argumentation?

Here is what he sent me -

	#include <stdio.h>
	
	void add ()
	{	
	     printf ("inside function: add. \n");
      	
	     return;
	}

	int main (void)
	{
	    /* call function add with some parameters */
	    add (5, 1);
     	
	    system ("PAUSE");
	
	    return (0);
	}

   How can this work, if not specifying any argument, is equivalent to 
specifying as void?
   However, one thing I was able observe was that it accepts any kind of 
arguments, and also any number of arguments, as against his theory of 
only accepting "int" types.
   I even tried compiling with "-Wall" option to see if any warnings are 
being thrown by the compiler, but found to my disappointment that there 
was none.
   Am I fundamentally going wrong in my understanding of functions?

_z33
-- 
I love TUX; well... that's an understatement :)