On 9/9/05, _z33 <[EMAIL PROTECTED]> wrote:
> Steve Graegert wrote:
> > On 9/9/05, _z33 <[EMAIL PROTECTED]> wrote:
> >
> >> I had a wierd doubt today morning. If a function's return type is not
> >>defined, "C" takes it as returning "int". Now, what does it do when I
> >>don't specify the arguments of the function. Something like this -
> >>
> >> void sampleFunc ()
> >> {
> >> /* ... */
> >> }
> >>
> >> Is this equivalent to saying,
> >>
> >> void sampleFunc (void)
> >> {
> >> /* ... */
> >> }
> >
> >
> > Yes, technically both are equivalent. The latter is the new style
> > while the former is the "old" style. But be aware: A function defined
> > using the old style does __not__ establish a prototype, but if a
> > previously declared prototype for that function exists, the parameter
> > declarations in the definition must exactly match those in the
> > prototype after the default argument promotions are applied to the
> > parameters in the definition.
> >
> > Conclusion: avoid mixing old style and prototype style
> > declarations/definition for a given function. It is allowed but not
> > recommended.
>
> I'm clear... but, now wondering why for two days a guy from an R&D
> dept of an MNC is arguing with me, saying that a function with empty
> argument specification implies having implicit "int" type arguments.
> (similar to the implicit assumption of return type of functions to "int"
> when none is specified explicitly).
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?
Regards
\Steve
--
Steve Graegert <[EMAIL PROTECTED]>
Software Consultancy {C/C++ && Java && .NET}
Mobile: +49 (176) 21248869
Office: +49 (9131) 7126409
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