7 Declarations [dcl.dcl]

An unnamed-namespace-definition behaves as if it were replaced by

inlineopt namespace unique  { /* empty body */ }
using namespace unique  ;
namespace unique  { namespace-body }

where inline appears if and only if it appears in the unnamed-namespace-definition, all occurrences of unique in a translation unit are replaced by the same identifier, and this identifier differs from all other identifiers in the entire program.94 [ Example:

namespace { int i; }            // unique  ::i
void f() { i++; }               // unique  ::i++

namespace A {
  namespace {
    int i;                      // A:: unique  ::i
    int j;                      // A:: unique  ::j
  }
  void g() { i++; }             // A:: unique  ::i++
}

using namespace A;
void h() {
  i++;                          // error: unique  ::i or A:: unique  ::i
  A::i++;                       // A:: unique  ::i
  j++;                          // A:: unique  ::j
}

 — end example ]

Members (including explicit specializations of templates ([temp.expl.spec])) of a namespace can be defined within that namespace. [ Example:

namespace X {
  void f() { /* ... */ }
}

 — end example ]

Members of a named namespace can also be defined outside that namespace by explicit qualification ([namespace.qual]) of the name being defined, provided that the entity being defined was already declared in the namespace and the definition appears after the point of declaration in a namespace that encloses the declaration's namespace. [ Example:

namespace Q {
  namespace V {
    void f();
  }
  void V::f() { /* ... */ }     // OK
  void V::g() { /* ... */ }     // error: g() is not yet a member of V
  namespace V {
    void g();
  }
}

namespace R {
  void Q::V::g() { /* ... */ }  // error: R doesn't enclose Q
}

 — end example ]

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 function95 the friend class or function is a member of the innermost enclosing namespace. The name of the friend is not found by unqualified lookup ([basic.lookup.unqual]) or by qualified lookup ([basic.lookup.qual]) until a matching declaration is provided in that namespace scope (either before or after the class definition granting friendship). If a friend function is called, its name may be found by the name lookup that considers functions from namespaces and classes associated with the types of the function arguments ([basic.lookup.argdep]). If the name in a friend declaration is neither qualified nor a template-id and the declaration is a function or an elaborated-type-specifier, the lookup to determine whether the entity has been previously declared shall not consider any scopes outside the innermost enclosing namespace. [ Note: The other forms of friend declarations cannot declare a new member of the innermost enclosing namespace and thus follow the usual lookup rules.  — end note ] [ Example:

// 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
}

using A::x;

void h() {
  A::f(x);
  A::X::f(x);                   // error: f is not a member of A::X
  A::X::Y::g();                 // error: g is not a member of A::X::Y
}

 — end example ]