Re: default function parameters

[Jarmo <SVisor@lycos.com>]

There is a diff in how C and C++ sees this.

For C add( ) is an function taking undefined argument(s), so you can 
send it whatever you want. C++ on other hand will see add( ) as add( 
void ), and would complain. To say that add( ) would be equal to add( 
int(s) ) is bogus thou.

In real life this does not matter. This is a completely academic 
question. Thou *I think* even the C compiler should have given a 
warning. (Note -Wall does not turn on all warnings, just almost all).

// Jarmo

-- 

>> 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