From: "David Abrahams" <[EMAIL PROTECTED]>
> "Paul Mensonides" <[EMAIL PROTECTED]> 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.
_______________________________________________
Unsubscribe & other changes: http://lists.boost.org/mailman/listinfo.cgi/boost
