Namespaces and friends (C++ only)

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, the friend class or function is a member of the innermost enclosing namespace.

The following is an example of this structure:

// f has not yet been defined
void z(int);
namespace A {
  class X {
    friend void f(X);  // A::f is a friend
    };
  // A::f is not visible here
  X x;
  void f(X) { /* definition */}  // f() is defined and known to be a friend
}

using A::x;

void z()
{
  A::f(x);    // OK
  A::X::f(x); // error: f is not a member of A::X
}

In this example, function f() can only be called through namespace A using the call A::f(s);. Attempting to call function f() through class X using the A::X::f(x); call results in a compiler error. Since the friend declaration first occurs in a non-local class, the friend function is a member of the innermost enclosing namespace and may only be accessed through that namespace.

Related information

• Friends (C++ only)