A qualified name is a name that appears on the right hand side of the scope resolution operator :: (see also qualified identifiers). A qualified name may refer to a.
- class member (including static and non-static functions, types, templates, etc)
- namespace member (including another namespace)
- enumerator
If there is nothing on the left hand side of the ::, the lookup considers only declarations made in the global namespace scope (or introduced into the global namespace by a using declaration). This makes it possible to refer to such names even if they were hidden by a local declaration:
#include <iostream>
int main()
{
struct std{};
std::cout << "fail\n"; // Error: unqualified lookup for 'std' finds the struct
::std::cout << "ok\n"; // OK: ::std finds the namespace std
}Before name lookup can be performed for the name on the right hand side of ::, lookup must be completed for the name on its left hand side (unless a decltype expression is used, or there is nothing on the left). This lookup, which may be qualified or unqualified, depending on whether there's another :: to the left of that name, considers only namespaces, class types, enumerations, and templates whose specializations are types:
struct A {
static int n;
};
int main() {
int A;
A::n = 42; // OK: unqualified lookup of A to the left of :: ignores the variable
A b; // error: unqualified lookup of A finds the variable A
}When a qualified name is used as a declarator, then lookup of all names used in the same declarator that follow that qualified name, but not the names that precede it, is performed as if qualified the same way:
class X { };
constexpr int number = 100;
class C {
class X { };
static const int number = 50;
static X arr[number];
};
X C::arr[number], brr[number]; // Error
// Every name in the declarator "C::arr[number]" after "C::arr"
// is looked up within C::, but the names before C::arr are unaffected,
// The names in the second declarator ("brr[number]") are also unaffected
// equivalent to:
// "::X C::arr[C::number], brr[::number]"
C::X C::arr[number], brr[number]; // Compiles, size of arr is 50, size of brr is 100If the name on the right hand side of :: is a destructor or pseudo-destructor (that is, the character ~ followed by an identifier), that identifier is looked up in the same scope as the name on the left hand side of ::
struct C { typedef int I; };
typedef int I1, I2;
extern int *p, *q;
struct A { ~A(); };
typedef A AB;
int main() {
p->C::I::~I(); // the name I after ~ is looked up in the same scope as I before ::
// (that is, within the scope of C, so it finds C::I)
q->I1::~I2(); // The name I2 is looked up in the same scope as I1
// that is, from the current scope, so it finds ::I2
AB x;
x.AB::~AB(); // The name AB after ~ is looked up in the same scope as AB before ::
// that is, from the current scope, so it finds ::AB
} EnumeratorsIf the lookup of the left-hand side name comes up with an enumeration (either scoped or unscoped), the lookup of the right-hand side must result in an enumerator that belongs that enumeration, otherwise the program is ill-formed. | (since C++11) |
Class members
If the lookup of the left hand side name comes up with a class/struct or union name, the name on the right hand side of :: is looked up in the scope of that class (and so may find a declaration of a member of that class or of its base), with the following exceptions.
- destructor name is looked up as described above (in the scope of the name to the left of ::)
- user-defined conversion function name is (TODO)
- names used in template arguments are looked up in the current scope (not in the scope of the template name)
- names in using-declarations also consider class/enum names that are hidden in current scope
If the right hand side of :: names the same class as the left hand side, the name designates the constructor of that class. Such qualified name can only be used in a declaration of a constructor and in the using-declaration for an inheriting constructor. In those lookups where function names are ignored (that is, when looking up a name on the left of ::, when looking up a name in elaborated type specifier, or base specifier), the same syntax resolves to the injected-class-name:
struct A { A(); };
struct B : A { B(); };
A::A() { } // A::A names a constructor, used in a declaration
B::B() { } // B::B names a constructor, used in a declaration
B::A ba; // B::A names the type A (looked up in the scope of B)
A::A a; // Error, A::A does not name a type
struct A::A a2; // OK: lookup in elaborated type specifier ignores functions
// so A::A simply names the class A as seen from within the scope of A
// (that is, the injected-class-name)Qualified name lookup can be used to access a class member that is hidden by a nested declaration or by a derived class. A call to a qualified member function is never virtual.
struct B { virtual void foo(); };
struct D : B { void foo() override; };
int main()
{
D x;
B& b = x;
b.foo(); // calls D::foo (virtual dispatch)
b.B::foo(); // calls B::foo (static dispatch)
}Namespace members
If the name on the left of :: refers to a namespace or if there is nothing on the left of :: (in which case it refers to the global namespace), the name that appears on the right hand side of :: is looked up in the scope of that namespace, except that.
- names used in template arguments are looked up in the current scope
namespace N {
template<typename T> struct foo {};
struct X {};
}
N::foo<X> x; // error: X is looked up as ::X, not as N::XQualified lookup within the scope of a namespace N first considers all declarations that are located in N and all declarations that are located in the inline namespace members of N (and, transitively, in their inline namespace members). If there are no declarations in that set then it considers declarations in all namespaces named by using-directives found in N and in all transitive inline namespace members of N. The rules are applied recursively:
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(); // AB is searched, AB::g found by lookup and is chosen AB::g(void)
// (A and B are not searched)
AB::f(1); // First, AB is searched, there is no f
// Then, A, B are searched
// A::f, B::f found by lookup (but Y is not searched so Y::f is not considered)
// overload resolution picks A::f(int)
AB::x++; // First, AB is searched, there is no x
// Then A, B are searched. There is no x
// Then Y and Z are searched. There is still no x: this is an error
AB::i++; // AB is searched, there is no i
// Then A, B are searched. A::i and B::i found by lookup: this is an error
AB::h(16.8); // First, AB is searched: there is no h
// Then A, B are searched. There is no h
// Then Y and Z are searched.
// lookup finds Y::h and Z::h. Overload resolution picks Z::h(double)
}It is allowed for the same declaration to be found more than once:
namespace A { int a; }
namespace B { using namespace A; }
namespace D { using A::a; }
namespace BD {
using namespace B;
using namespace D;
}
void g()
{
BD::a++; // OK: finds the same A::a through B and through D
}
Please login to continue.