Qualified name lookup

Qualified name lookup

Unqualified identifiers(omitted)

Besides suitably declared identifiers, the following can be used in expressions in the same role:

an overloaded operator name in function notation, such as operator+ or operator new
a user-defined conversion function name, such as operator bool
a user-defined literal operator name, such as operator "" _km
a template name followed by its argument list, such as MyTemplate<int>
the character ~ followed by a class name, such as ~MyClass
the character ~ followed by a decltype specifier, such as ~decltype(str)

Together with identifiers they are known as unqualified id-expressions.

Qualified identifiers

A qualified id-expression is an unqualified id-expression prepended by a scope resolution operator ::, and optionally, a sequence of enumeration, (since C++11)class or namespace names or decltype expressions (since C++11) separated by scope resolution operators.

std::string::npos

::tolower

std::cout

boost::signals2::connection

###Qualified name lookup
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
* [enumeration](#enumerations)
* [**class member**](#class)(including static and non-static function,types,templates.etc)
* [**namespace member**](#namespace)(including another namespace)

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).

Example hided std namespace

#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,global namespace

}

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:

在name lookup执行之前，要检查name的完整性，::右边的name在左边的scope中,则是qualified,否则为unqualified。

::右边的名字必须在::左边的namespace or class types or enumeration or templates whose specializations are types 中。

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]"

  // 这与声明 int *a,b;一样，a是一个int*类型，而b是一个int类型

  // 因为在X类型中找不到

C::X C::arr[number], brr[number]; // Compiles, size of arr is 50, size of brr is 100

destructor的lookup

If 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

}

###Enumerations
Enumerators
If 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.

###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 argument** are looked up in the current scope(not int the scope of template name)
* names in using-declarations also consider class/enum names that are hidden in current scope

编译器中constructor的产生

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
1. 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 int the current scope.
```cpp
namespace N {
template struct foo {};
struct X {};
}
N::foo x; // error: X is looked up as ::X, not as N::X
```

在namespace N内的qualified lookup首先会考虑到位于N内的和N内的inline namespace 成员(具有传递性)。

2. Qualified 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).

之后才会考虑到使用use-derectives引入的namespace和inline namespace(使用use-derectives引入的namespace)

3. 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++;  // ERROR   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++;  // ERROR    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

}