Introduces a name that is defined elsewhere into the declarative region where this using-declaration appears.

using typename(optional) nested-name-specifier unqualified-id ; (until C++17)
using declarator-list ; (since C++17)
nested-name-specifier - a sequence of names and scope resolution operators ::, ending with a scope resolution operator. A single :: refers to the global namespace.
unqualified-id - an id-expression
typename - the keyword typename may be used as necessary to resolve dependent names, when the using-declaration introduces a member type from a base class into a class template
declarator-list - comma-separated list of one or more declarators of the typename(optional) nested-name-specifier unqualified-id. Some or all of the declarators may be followed by an ellipsis ... to indicate pack expansion


Using-declarations can be used to introduce namespace members into other namespaces and block scopes, or to introduce base class members into derived class definitions.

A using-declaration with more than one using-declarator is equivalent to a corresponding sequence of using-declarations with one using-declarator.

(since C++17)

In namespace and block scope

Using-declaration introduces a member of another namespace into current namespace or block scope.

#include <iostream>
#include <string>
using std::string;
int main()
    string str = "Example";
    using std::cout;
    cout << str;

See namespace for details.

In class definition

Using-declaration introduces a member of a base class into the derived class definition, such as to expose a protected member of base as public member of derived. In this case, nested-name-specifier must name a base class of the one being defined. If the name is the name of an overloaded member function of the base class, all base class member functions with that name are introduced. If the derived class already has a member with the same name, parameter list, and qualifications, the derived class member hides or overrides (doesn't conflict with) the member that is introduced from the base class.

#include <iostream>
struct B {
    virtual void f(int) { std::cout << "B::f\n"; }
    void g(char)        { std::cout << "B::g\n"; }
    void h(int)         { std::cout << "B::h\n"; }
    int m; // B::m is protected
    typedef int value_type;
struct D : B {
    using B::m; // D::m is public
    using B::value_type; // D::value_type is public
    using B::f;
    void f(int) { std::cout << "D::f\n"; } // D::f(int) overrides B::f(int)
    using B::g;
    void g(int) { std::cout << "D::g\n"; } // both g(int) and g(char) are visible
                                           // as members of D
    using B::h;
    void h(int) { std::cout << "D::h\n"; } // D::h(int) hides B::h(int)
int main()
    D d;
    B& b = d;
//    b.m = 2; // error, B::m is protected
    d.m = 1; // protected B::m is accessible as public D::m
    b.f(1); // calls derived f()
    d.f(1); // calls derived f()
    d.g(1); // calls derived g(int)
    d.g('a'); // calls base g(char)
    b.h(1); // calls base h()
    d.h(1); // calls derived h()



Inheriting constructors

If the using-declaration refers to a constructor of a direct base of the class being defined (e.g. using Base::Base;), constructors of that base class are inherited, according to the following rules:

1) A set of candidate inheriting constructors is composed of a) All non-template constructors of the base class (after omitting ellipsis parameters, if any) (since C++14) b) For each constructor with default arguments or the ellipsis parameter, all constructor signatures that are formed by dropping the ellipsis and omitting default arguments from the ends of argument lists one by one c) All constructor templates of the base class (after omitting ellipsis parameters, if any) (since C++14) d) For each constructor template with default arguments or the ellipsis, all constructor signatures that are formed by dropping the ellipsis and omitting default arguments from the ends of argument lists one by one 2) All candidate inherited constructors that aren't the default constructor or the copy/move constructor and whose signatures do not match user-defined constructors in the derived class, are implicitly declared in the derived class. The default parameters are not inherited:
struct B1 {
struct D1 : B1 {
    using B1::B1;
// The set of candidate inherited constructors is 
// 1. B1(const B1&)
// 2. B1(B1&&)
// 3. B1(int)
// D1 has the following constructors:
// 1. D1() = delete
// 2. D1(const D1&) 
// 3. D1(D1&&)
// 4. D1(int) <- inherited
struct B2 {
    B2(int = 13, int = 42);
struct D2 : B2 {
    using B2::B2;
// The set of candidate inherited constructors is
// 1. B2(const B2&)
// 2. B2(B2&&)
// 3. B2(int = 13, int = 42)
// 4. B2(int = 13)
// 5. B2()
// D2 has the following constructors:
// 1. D2()
// 2. D2(const D2&)
// 3. D2(D2&&)
// 4. D2(int, int) <- inherited
// 5. D2(int) <- inherited

The inherited constructors are equivalent to user-defined constructors with an empty body and with a member initializer list consisting of a single nested-name-specifier, which forwards all of its arguments to the base class constructor.

It has the same access as the corresponding base constructor. It is constexpr if the user-defined constructor would have satisfied constexpr constructor requirements. It is deleted if the corresponding base constructor is deleted or if a defaulted default constructor would be deleted (except that the construction of the base whose constructor is being inherited doesn't count). An inheriting constructor cannot be explicitly instantiated or explicitly specialized.

If two using-declarations inherit the constructor with the same signature (from two direct base classes), the program is ill-formed.

(since C++11)
(until C++17)

If the using-declaration refers to a constructor of a direct base of the class being defined (e.g. using Base::Base;), all constructors of that base (ignoring member access) are made visible to overload resolution when initializing the derived class.

If overload resolution selects an inherited constructor, it is accessible if it would be accessible when used to construct an object of the corresponding base class: the accessibility of the using-declaration that introduced it is ignored.

If overload resolution selects one of the inherited constructors when initializing an object of such derived class, then the Base subobject from which the constructor was inherited is initialized using the inherited constructor, and all other bases and members of Derived are initialized as if by the defaulted default constructor (default member initializers are used if provided, otherwise default initialization takes place). The entire initialization is treated as a single function call: initialization of the parameters of the inherited constructor is sequenced-before initialization of any base or member of the derived object.

struct B1 {  B1(int, ...) { } };
struct B2 {  B2(double)   { } };
int get();
struct D1 : B1 {
  using B1::B1;  // inherits B1(int, ...)
  int x;
  int y = get();
void test() {
  D1 d(2, 3, 4); // OK: B1 is initialized by calling B1(2, 3, 4),
                 // then d.x is default-initialized (no initialization is performed),
                 // then d.y is initialized by calling get()
  D1 e;          // Error: D1 has no default constructor
struct D2 : B2 {
  using B2::B2; // inherits B2(double)
  B1 b;
D2 f(1.0);       // error: B1 has no default constructor
struct W { W(int); };
struct X : virtual W {
 using W::W;   // inherits W(int)
 X() = delete;
struct Y : X {
 using X::X;
struct Z : Y, virtual W {
  using Y::Y;
Z z(0); // OK: initialization of Y does not invoke default constructor of X

If the constructor was inherited from multiple base class subobjects of type B, the program is ill-formed, similar to multiply-inherited non-static member functions:

struct A { A(int); };
struct B : A { using A::A; };
struct C1 : B { using B::B; };
struct C2 : B { using B::B; };
struct D1 : C1, C2 {
  using C1::C1;
  using C2::C2;
D1 d1(0); // ill-formed: constructor inherited from different B base subobjects
struct V1 : virtual B { using B::B; };
struct V2 : virtual B { using B::B; };
struct D2 : V1, V2 {
  using V1::V1;
  using V2::V2;
D2 d2(0); // OK: there is only one B subobject.
          // This initializes the virtual B base class,
          //  which initializes the A base class
          // then initializes the V1 and V2 base classes
          //  as if by a defaulted default constructor

As with using-declarations for any other non-static member functions, if an inherited constructor matches the signature of one of the constructors of Derived, it is hidden from lookup by the version found in Derived. If one of the inherited constructors of Base happens to have the signature that matches a copy/move constructor of the Derived, it does not prevent implicit generation of Derived copy/move constructor (which then hides the inherited version, similar to using operator=).

struct B1 {   B1(int); };
struct B2 {   B2(int); };
struct D2 : B1, B2 {
  using B1::B1;
  using B2::B2;
  D2(int);   // OK: D2::D2(int) hides both B1::B1(int) and B2::B2(int)
D2 d2(0);    // calls D2::D2(int)
(since C++17)


Only the name explicitly mentioned in the using-declaration is transferred into the declarative scope: in particular, enumerators are not transferred when the enumeration type name is using-declared.

A using-declaration cannot refer to a namespace, to a scoped enumerator, to a destructor of a base class or to a specialization of a member template for a user-defined conversion function.

Using-declaration can refer to a member template, but cannot refer to a member template specialization (template-id is not permitted).

struct B { template<class T> void f(); };
struct D : B {
      using B::f;      // OK: names a template
//    using B::f<int>; // Error: names a template specialization
      void g() { f<int>(); }

Using-declaration also can't refer to a dependent member template (the template disambiguator for dependent names is not permitted).

template<class X> struct B { template<class T> void f(); };
template<class Y> struct D : B<Y> {
//  using B<Y>::template f; // Error: disambiguator not allowed
  using B<Y>::f;            // compiles, but declares f as non-type non-template
  void g() { f<int>(); }    // Error here

If a using-declaration brings the base class assignment operator into derived class, whose signature happens to match the derived class's copy-assignment or move-assignment operator, that operator is hidden by the implicitly-declared copy/move assignment operator of the derived class. Same applies to a using-declaration that inherits a base class constructor that happens to match the derived class copy/move constructor (since C++17).

Pack expansions in using-declarations make it possible to form a class that exposes overloaded members of variadic bases without recursion:

template <typename... Ts>
struct Overloader : Ts... {
    using Ts::operator()...; // exposes operator() from every base
template <typename... T>
Overloader(T...) -> Overloader<T...>; // C++17 deduction guide
int main() {
    auto o = Overloader{ [] (auto const& a) {std::cout << a;},
                         [] (float f) {std::cout << std::setprecision(3) << f;} };
(since C++17)

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