Why does a C++ friend class need a forward declaration only in other namespaces?

Question

Suppose I have a class F that should be friend to the classes G (in the global namespace) and C (in namespace A).

• to be friend to A::C, F must be forward declared.
• to be friend to G, no forward declaration of F is necessary.
• likewise, a class A::BF can be friend to A::C without forward declaration

The following code illustrates this and compiles with GCC 4.5, VC++ 10 and at least with one other compiler.

class G {     friend class F;     int g; };  // without this forward declaration, F can't be friend to A::C class F;  namespace A {  class C {     friend class ::F;     friend class BF;     int c; };  class BF { public:     BF() { c.c = 2; } private:     C c; };  } // namespace A  class F { public:     F() { g.g = 3; c.c = 2; } private:     G g;     A::C c; };  int main() {     F f; } 

To me this seems inconsistent. Is there a reason for this or is it just a design decision of the standard?

Answer 1

C++ Standard ISO/IEC 14882:2003(E)

7.3.1.2 Namespace member definitions

Paragraph 3

Every name first declared in a namespace is a member of that namespace. If a friend declaration in a non-local class first declares a class or function (this implies that the name of the class or function is unqualified) the friend class or function is a member of the innermost enclosing namespace.

// Assume f and g have not yet been defined. void h(int); template <class T> void f2(T); namespace A {    class X {    friend void f(X);  //  A::f(X) is a friend       class Y {          friend void g();  //  A::g is a friend          friend void h(int);  //  A::h is a friend          //  ::h not considered          friend void f2<>(int);  //  ::f2<>(int) is a friend       };    };    //  A::f, A::g and A::h are not visible here    X x;    void g() { f(x); }  // definition of A::g    void f(X) { /* ... */}  // definition of A::f    void h(int) { /* ... */ }  // definition of A::h    //  A::f, A::g and A::h are visible here and known to be friends } 

Your friend class BF; is a declaration of A::BF in namespace A rather than global namespace. You need the global prior declaration to avoid this new declaration.

Answer 2

Let's take into account these 3 code lines from your sample:

1. friend class F; // it creates "friend declaration", (that's not the same as ordinary forward declaration  2. class F; // without this forward declaration, F can't be friend to A::C <-- this is ordinary forward declaration  3. friend class ::F; // this is qualified lookup (because of ::), so it needs previous declaration, which you provide in line 2. 

C++ standard in paragraph 7.3.1.2, point 3 ( Namespace member definitions) says:

The friend declaration does not by itself make the name visible to unqualified lookup (3.4.1) or qualified lookup (3.4.3). [ Note: The name of the friend will be visible in its namespace if a matching declaration is provided at namespace scope (either before or after the class definition granting friendship). —end note ]

And line 2 follows exactly what standard requires.

All confusion is because "friend declaration" is weak, you need to provide solid forward declaration for further usage.