6 Basics [basic]

6.5 Name lookup [basic.lookup]

6.5.3 Qualified name lookup [basic.lookup.qual]

1
#
The name of a class or namespace member or enumerator can be referred to after the ​::​ scope resolution operator ([expr.prim.id.qual]) applied to a nested-name-specifier that denotes its class, namespace, or enumeration.
If a ​::​ scope resolution operator in a nested-name-specifier is not preceded by a decltype-specifier, lookup of the name preceding that ​::​ considers only namespaces, types, and templates whose specializations are types.
If the name found does not designate a namespace or a class, enumeration, or dependent type, the program is ill-formed.
[Example
: 
class A {
public:
  static int n;
};
int main() {
  int A;
  A::n = 42;        // OK
  A b;              // error: A does not name a type
}
— end example
]
2
#
[Note
:
Multiply qualified names, such as N1​::​N2​::​N3​::​n, can be used to refer to members of nested classes ([class.nest]) or members of nested namespaces.
— end note
]
3
#
In a declaration in which the declarator-id is a qualified-id, names used before the qualified-id being declared are looked up in the defining namespace scope; names following the qualified-id are looked up in the scope of the member's class or namespace.
[Example
: 
class X { };
class C {
  class X { };
  static const int number = 50;
  static X arr[number];
};
X C::arr[number];   // error:
                    // equivalent to ​::​X C​::​arr[C​::​number];
                    // and not to C​::​X C​::​arr[C​::​number];
— end example
]
4
#
A name prefixed by the unary scope operator ​::​ ([expr.prim.id.qual]) is looked up in global scope, in the translation unit where it is used.
The name shall be declared in global namespace scope or shall be a name whose declaration is visible in global scope because of a using-directive ([namespace.qual]).
The use of ​::​ allows a global name to be referred to even if its identifier has been hidden.
5
#
A name prefixed by a nested-name-specifier that nominates an enumeration type shall represent an enumerator of that enumeration.
6
#
In a qualified-id of the form:
nested-name-specifier type-name :: ~ type-name
the second type-name is looked up in the same scope as the first.
[Example
: 
struct C {
  typedef int I;
};
typedef int I1, I2;
extern int* p;
extern int* q;
p->C::I::~I();      // I is looked up in the scope of C
q->I1::~I2();       // I2 is looked up in the scope of the postfix-expression

struct A {
  ~A();
};
typedef A AB;
int main() {
  AB* p;
  p->AB::~AB();     // explicitly calls the destructor for A
}
— end example
]
[Note
:
[basic.lookup.classref] describes how name lookup proceeds after the . and -> operators.
— end note
]

6.5.3.1 Class members [class.qual]

1
#
If the nested-name-specifier of a qualified-id nominates a class, the name specified after the nested-name-specifier is looked up in the scope of the class ([class.member.lookup]), except for the cases listed below.
The name shall represent one or more members of that class or of one of its base classes ([class.derived]).
[Note
:
A class member can be referred to using a qualified-id at any point in its potential scope ([basic.scope.class]).
— end note
]
The exceptions to the name lookup rule above are the following:
2
#
In a lookup in which function names are not ignored26 and the nested-name-specifier nominates a class C: the name is instead considered to name the constructor of class C.
[Note
:
For example, the constructor is not an acceptable lookup result in an elaborated-type-specifier so the constructor would not be used in place of the injected-class-name.
— end note
]
Such a constructor name shall be used only in the declarator-id of a declaration that names a constructor or in a using-declaration.
[Example
: 
struct A { A(); };
struct B: public A { B(); };

A::A() { }
B::B() { }

B::A ba;            // object of type A
A::A a;             // error: A​::​A is not a type name
struct A::A a2;     // object of type A
— end example
]
3
#
A class member name hidden by a name in a nested declarative region or by the name of a derived class member can still be found if qualified by the name of its class followed by the ​::​ operator.
26)
Lookups in which function names are ignored include names appearing in a nested-name-specifier, an elaborated-type-specifier, or a base-specifier.
⮥

6.5.3.2 Namespace members [namespace.qual]

1
#
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.
2
#
For a namespace X and name m, the namespace-qualified lookup set is defined as follows: Let be the set of all declarations of m in X and the inline namespace set of X ([namespace.def]) whose potential scope ([basic.scope.namespace]) would include the namespace in which m is declared at the location of the nested-name-specifier.
If is not empty, is ; otherwise, is the union of for all namespaces nominated by using-directives in X and its inline namespace set.
3
#
Given X​::​m (where X is a user-declared namespace), or given ​::​m (where X is the global namespace), if is the empty set, the program is ill-formed.
Otherwise, if has exactly one member, or if the context of the reference is a using-declaration, 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 , 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  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  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  and
                // overload resolution chooses Z​::​h(double)
}
— end example
]
4
#
[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 
}

namespace D {
  using A::a;
}

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

void g()
{
  BD::a++;          // OK: S is 
}
— end example
]
— end note
]
5
#
[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
]
6
#
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 introduce either 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
]
7
#
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){ }  // error: 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
]