6 Basics [basic]

6.5 Name lookup [basic.lookup]

6.5.5 Qualified name lookup [basic.lookup.qual]

6.5.5.1 General [basic.lookup.qual.general]

1
#
Lookup of an identifier followed by a ​::​ scope resolution operator considers only namespaces, types, and templates whose specializations are types.
If a name, template-id, or computed-type-specifier is followed by a ​::​, it shall designate a namespace, class, enumeration, or dependent type, and the ​::​ is never interpreted as a complete nested-name-specifier.
[Example 1: class A { public: static int n; }; int main() { int A; A::n = 42; // OK A b; // error: A does not name a type } template<int> struct B : A {}; namespace N { template<int> void B(); int f() { return B<0>::n; // error: N​::​B<0> is not a type } } — end example]
2
#
A member-qualified name is the (unique) component name ([expr.prim.id.unqual]), if any, of in the id-expression of a class member access expression ([expr.ref]).
The lookup context of a member-qualified name is the type of its associated object expression (considered dependent if the object expression is type-dependent).
The lookup context of any other qualified name is the type, template, or namespace nominated by the preceding nested-name-specifier.
[Note 1: 
When parsing a class member access, the name following the -> or . is a qualified name even though it is not yet known of which kind.
— end note]
[Example 2: 
In N::C::m.Base::f() Base is a member-qualified name; the other qualified names are C, m, and f.
— end example]
3
#
Qualified name lookup in a class, namespace, or enumeration performs a search of the scope associated with it ([class.member.lookup]) except as specified below.
Unless otherwise specified, a qualified name undergoes qualified name lookup in its lookup context from the point where it appears unless the lookup context either is dependent and is not the current instantiation ([temp.dep.type]) or is not a class or class template.
If nothing is found by qualified lookup for a member-qualified name that is the terminal name ([expr.prim.id.unqual]) of a nested-name-specifier and is not dependent, it undergoes unqualified lookup.
[Note 2: 
During lookup for a template specialization, no names are dependent.
— end note]
[Example 3: int f(); struct A { int B, C; template<int> using D = void; using T = void; void f(); }; using B = A; template<int> using C = A; template<int> using D = A; template<int> using X = A; template<class T> void g(T *p) { // as instantiated for g<A>: p->X<0>::f(); // error: A​::​X not found in ((p->X) < 0) > ​::​f() p->template X<0>::f(); // OK, ​::​X found in definition context p->B::f(); // OK, non-type A​::​B ignored p->template C<0>::f(); // error: A​::​C is not a template p->template D<0>::f(); // error: A​::​D<0> is not a class type p->T::f(); // error: A​::​T is not a class type } template void g(A*); — end example]
4
#
If a qualified name Q follows a ~:
  • (4.1)
    If Q is a member-qualified name, it undergoes unqualified lookup as well as qualified lookup.
  • (4.2)
    Otherwise, its nested-name-specifier N shall nominate a type.
    If N has another nested-name-specifier S, Q is looked up as if its lookup context were that nominated by S.
  • (4.3)
    Otherwise, if the terminal name of N is a member-qualified name M, Q is looked up as if ~Q appeared in place of M (as above).
  • (4.4)
    Otherwise, Q undergoes unqualified lookup.
  • (4.5)
    Each lookup for Q considers only types (if Q is not followed by a <) and templates whose specializations are types.
    If it finds nothing or is ambiguous, it is discarded.
  • (4.6)
    The type-name that is or contains Q shall refer to its (original) lookup context (ignoring cv-qualification) under the interpretation established by at least one (successful) lookup performed.
[Example 4: struct C { typedef int I; }; typedef int I1, I2; extern int* p; extern int* q; void f() { p->C::I::~I(); // I is looked up in the scope of C q->I1::~I2(); // I2 is found by unqualified lookup } struct A { ~A(); }; typedef A AB; int main() { AB* p; p->AB::~AB(); // explicitly calls the destructor for A } — end example]

6.5.5.2 Class members [class.qual]

1
#
In a lookup for a qualified name N whose lookup context is a class C in which function names are not ignored,18 N is instead considered to name the constructor of class C.
Such a constructor name shall be used only in the declarator-id of a (friend) declaration of a constructor or in a using-declaration.
[Example 1: 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]
18)18)
Lookups in which function names are ignored include names appearing in a nested-name-specifier, an elaborated-type-specifier, or a base-specifier.

6.5.5.3 Namespace members [namespace.qual]

1
#
Qualified name lookup in a namespace N additionally searches every element of the inline namespace set of N ([namespace.def]).
If nothing is found, the results of the lookup are the results of qualified name lookup in each namespace nominated by a using-directive that precedes the point of the lookup and inhabits N or an element of N's inline namespace set.
[Note 1: 
If a using-directive refers to a namespace that has already been considered, it does not affect the result.
— end note]
[Example 1: 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]
2
#
[Note 2: 
The same declaration found more than once is not an ambiguity (because it is still a unique declaration).
[Example 2: 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]
3
#
[Example 3: 
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]
4
#
[Note 3: 
Class and enumeration declarations are not discarded because of other declarations found in other searches.
— end note]
[Example 4: 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]