6 Basic concepts [basic]

6.4 Name lookup [basic.lookup]

6.4.3 Qualified name lookup [basic.lookup.qual]

6.4.3.2 Namespace members [namespace.qual]

If the nested-name-specifier of a qualified-id nominates a namespace (including the case where the nested-name-specifier is ​::​, i.e., nominating the global namespace), the name specified after the nested-name-specifier is looked up in the scope of the namespace. The names in a template-argument of a template-id are looked up in the context in which the entire postfix-expression occurs.

For a namespace X and name m, the namespace-qualified lookup set S(X,m) is defined as follows: Let S(X,m) be the set of all declarations of m in X and the inline namespace set of X. If S(X,m) is not empty, S(X,m) is S(X,m); otherwise, S(X,m) is the union of S(Ni,m) for all namespaces Ni nominated by using-directives in X and its inline namespace set.

Given X​::​m (where X is a user-declared namespace), or given ​::​m (where X is the global namespace), if S(X,m) is the empty set, the program is ill-formed. Otherwise, if S(X,m) has exactly one member, or if the context of the reference is a using-declaration, S(X,m) is the required set of declarations of m. Otherwise if the use of m is not one that allows a unique declaration to be chosen from S(X,m), the program is ill-formed. [Example:

int x;
namespace Y {
  void f(float);
  void h(int);
}

namespace Z {
  void h(double);
}

namespace A {
  using namespace Y;
  void f(int);
  void g(int);
  int i;
}

namespace B {
  using namespace Z;
  void f(char);
  int i;
}

namespace AB {
  using namespace A;
  using namespace B;
  void g();
}

void h()
{
  AB::g();      // g is declared directly in AB, therefore S is { AB​::​g() } and AB​::​g() is chosen

  AB::f(1);     // f is not declared directly in AB so the rules are applied recursively to A and B;
                // namespace Y is not searched and Y​::​f(float) is not considered;
                // S is {A::f(int),B::f(char)} and overload resolution chooses A​::​f(int)

  AB::f('c');   // as above but resolution chooses B​::​f(char)

  AB::x++;      // x is not declared directly in AB, and is not declared in A or B, so the rules
                // are applied recursively to Y and Z, S is { } so the program is ill-formed

  AB::i++;      // i is not declared directly in AB so the rules are applied recursively to A and B,
                // S is {A::i,B::i} so the use is ambiguous and the program is ill-formed

  AB::h(16.8);  // h is not declared directly in AB and not declared directly in A or B so the rules
                // are applied recursively to Y and Z, S is {Y::h(int),Z::h(double)} and
                // overload resolution chooses Z​::​h(double)
}

end example]

[Note: The same declaration found more than once is not an ambiguity (because it is still a unique declaration). [Example:

namespace A {
  int a;
}

namespace B {
  using namespace A;
}

namespace C {
  using namespace A;
}

namespace BC {
  using namespace B;
  using namespace C;
}

void f()
{
  BC::a++;          // OK: S is {A::a,A::a}
}

namespace D {
  using A::a;
}

namespace BD {
  using namespace B;
  using namespace D;
}

void g()
{
  BD::a++;          // OK: S is {A::a,A::a}
}

end example] end note]

[Example: Because each referenced namespace is searched at most once, the following is well-defined:

namespace B {
  int b;
}

namespace A {
  using namespace B;
  int a;
}

namespace B {
  using namespace A;
}

void f()
{
  A::a++;           // OK: a declared directly in A, S is { A​::​a }
  B::a++;           // OK: both A and B searched (once), S is { A​::​a }
  A::b++;           // OK: both A and B searched (once), S is { B​::​b }
  B::b++;           // OK: b declared directly in B, S is { B​::​b }
}

end example]

During the lookup of a qualified namespace member name, if the lookup finds more than one declaration of the member, and if one declaration introduces a class name or enumeration name and the other declarations either introduce the same variable, the same enumerator or a set of functions, the non-type name hides the class or enumeration name if and only if the declarations are from the same namespace; otherwise (the declarations are from different namespaces), the program is ill-formed. [Example:

namespace A {
  struct x { };
  int x;
  int y;
}

namespace B {
  struct y { };
}

namespace C {
  using namespace A;
  using namespace B;
  int i = C::x;     // OK, A​::​x (of type int)
  int j = C::y;     // ambiguous, A​::​y or B​::​y
}

end example]

In a declaration for a namespace member in which the declarator-id is a qualified-id, given that the qualified-id for the namespace member has the form

nested-name-specifier unqualified-id

the unqualified-id shall name a member of the namespace designated by the nested-name-specifier or of an element of the inline namespace set of that namespace. [Example:

namespace A {
  namespace B {
    void f1(int);
  }
  using namespace B;
}
void A::f1(int){ }  // ill-formed, f1 is not a member of A

end example] However, in such namespace member declarations, the nested-name-specifier may rely on using-directives to implicitly provide the initial part of the nested-name-specifier. [Example:

namespace A {
  namespace B {
    void f1(int);
  }
}

namespace C {
  namespace D {
    void f1(int);
  }
}

using namespace A;
using namespace C::D;
void B::f1(int){ }  // OK, defines A​::​B​::​f1(int)

end example]