In article <JASON.94May6165247@deneb.cygnus.com> jason@cygnus.com (Jason Merrill) writes:
>Is this code well-formed?

 Who can tell? There is no type 'member function'.
There are only pointers to them. But that doesnt automatically rule
out a typedef, since a typedef is just a syntax macro -- its supposed
to be an alias for a "type" but "type" isnt so well defined.

>If not, how do you define a typedef for declaring methods?

 Maybe you cant, maybe you can. Its been discussed.
I dont recall a resolution (which doesnt mean there wasnt one :-)
>
>struct A {
>  typedef void (A::fn)();
>};
>
>struct B: public A {
>  fn f;
>};
>
>void B::f () { }

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Is this code well-formed?  If so, why?  Does the method basetype get
coerced silently to the derived type?  Please provide a citation from the
latest WP.  If not, how do you define a typedef for declaring methods?

struct A {
  typedef void (A::fn)();
};

struct B: public A {
  fn f;
};

void B::f () { }

Jason

In article <JASON.94May6165247@deneb.cygnus.com> jason@cygnus.com (Jason Merrill) writes:
>Is this code well-formed?  If so, why?  Does the method basetype get
>coerced silently to the derived type?  Please provide a citation from the
>latest WP.  If not, how do you define a typedef for declaring methods?
>
>struct A {
>  typedef void (A::fn)();
>};
>
>struct B: public A {
>  fn f;
>};
>
>void B::f () { }

Jason,

I really wish that I could cite chapter and verse on this one, but all I
can tell you is what I believe.

I believe that it is not valid to declared *any* member using a class-
qualified name.  Thus, the typedef in your code above is invalid.

This is not very different from:

 struct B;

 struct A {
  int B::member;
 };

which, I believe, is also invalid.

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