Re: [boost] Fun, only handled by vc6/7!

From: Peter Dimov (pdimov_at_[hidden])
Date: 2003-01-13 09:43:04

From: "David Abrahams" <dave_at_[hidden]>
> "Paul Mensonides" <pmenso57_at_[hidden]> writes:
>
> >> template <class T> struct voidify { typedef void type; };
> >> template <class T> struct Y {};
> >> struct X
> >> {
> >>     template <class T>
> >>     operator Y<T> (typename voidify<T>::type) const { return Y<T>(); }
> >> };
> >
> > Is this even legal?  I.e. for a user-defined conversion operator to have
any
> > arguments at all?
>
> Look twice; the argument is void.

Doesn't matter. f(T) has one argument for any T. f(void) (void is a keyword
here, not a type expression) has zero arguments.