Should:

  template<class A> struct T {
   virtual void f( A ) = 0;
  }

  struct S : T<void> {
   void f() { }
  };

 have S::f override T::f?  I would think so since, after all, f()
 is equivalent to f(void).  The above makes sense for, say, int;
 but what about void?  My compiler complains that S::f hides
 T::f, but likes:

  template<class R> struct T {
   virtual R f() = 0;
  }

  struct S : T<void> {
   void f() { }
  };

 just fine.  Does the WP address this?
--
 - Paul J. Lucas   Paul.J.Lucas@att.com
   AT&T Bell Laboratories http://eec.psu.edu/~scott/pjl/
   Naperville, IL  #include <stddisclaimer.h>

In article <pjl.798130465@graceland.att.com> pjl@graceland.att.com (Paul J. Lucas) writes:
> Should:
>
>  template<class A> struct T {
>   virtual void f( A ) = 0;
>  }
>
>  struct S : T<void> {
>   void f() { }
>  };
>
> have S::f override T::f?  I would think so since, after all, f()
> is equivalent to f(void).  The above makes sense for, say, int;
> but what about void?

The example above is ill-formed when A is void.  A parameter cannot have
type "void".  Only the syntactic construct (void) is equivalent to ().
Other names for void (e.g., typedefs, etc.) cannot be used for this
purpose.


> My compiler complains that S::f hides
> T::f, but likes:
>
>  template<class R> struct T {
>   virtual R f() = 0;
>  }
>
>  struct S : T<void> {
>   void f() { }
>  };
>
> just fine.  Does the WP address this?

The WP does.  8.3.5p2.

John Spicer
Edison Design Group
jhs@edg.com