Re: Difference in compilation success with friend declaration

From:

=?iso-8859-1?q?Daniel_Kr=FCgler?= <daniel.kruegler@googlemail.com>

Newsgroups:

comp.lang.c++.moderated

Date:

Fri, 27 Apr 2007 15:54:59 CST

Message-ID:

<1177693669.599122.10010@c18g2000prb.googlegroups.com>

John Moeller schrieb:

If not, would I have to declare C public? Or is there
some other way to declare the friend that would work?

Regarding a current workaround I propose the
following replacement of your implementation:

Note that I introduced a publicly accessable
class CP which has a interface that can only
be used by the dedicated do_foo<CP>() function
similarily to your original question. If you
accept this change, you could even save class
CP and make C similarily publicly available.

//---------------------------------------
#include <iostream>
#include <ostream>

template <typename X>
void do_foo() {}

template <typename X>
struct A {
   void foo() {
     do_foo<X>();
   }
};

struct CP;

class B {
   struct C {};
   friend class CP;
public:
   A<CP> a;
};

class CP {
  B::C data;
  enum Janitor { Ticket };
  friend void do_foo<CP>();
public:
  static void foo(Janitor) {
    std::cout << "Different output...\n";
  }
};

template<>
void do_foo<CP>()
{
  CP::foo(CP::Ticket);
}

int main()
{
   B().a.foo();
}
//---------------------------------------

Greetings from Bremen,

Daniel Kr|gler

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