On 9/9/05, Jarmo <[EMAIL PROTECTED]> wrote:
> 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.
Exactly. A point I have missed to mention. Thanks.
> Thou *I think* even the C compiler should have given a
> warning. (Note -Wall does not turn on all warnings, just almost all).
try -ansi combined with -pedantic.
> >> 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
>
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