Difference between revisions of "cpp/language/friend"

Revision as of 02:44, 7 December 2017

The friend declaration appears in a class body and grants a function or another class access to private and protected members of the class where the friend declaration appears.

Syntax

Description

class Y {
    int data; // private member
    // the non-member function operator<< will have access to Y's private members
    friend std::ostream& operator<<(std::ostream& out, const Y& o);
    friend char* X::foo(int); // members of other classes can be friends too
    friend X::X(char), X::~X(); // constructors and destructors can be friends
};
// friend declaration does not declare a member function
// this operator<< still needs to be defined, as a non-member
std::ostream& operator<<(std::ostream& out, const Y& y)
{
    return out << y.data; // can access private member Y::data
}

class X {
    int a;
    friend void friend_set(X& p, int i) {
        p.a = i; // this is a non-member function
    }
 public:
    void member_set(int i) {
        a = i; // this is a member function
    }
};

class Y {};
class A {
    int data; // private data member
    class B { }; // private nested type
    enum { a = 100 }; // private enumerator
    friend class X; // friend class forward declaration (elaborated class specifier)
    friend Y; // friend class declaration (simple type specifier) (since c++11)
};
 
class X : A::B { // OK: A::B accessible to friend
    A::B mx; // OK: A::B accessible to member of friend
    class Y {
        A::B my; // OK: A::B accessible to nested member of friend
    };
    int v[A::a]; // OK: A::a accessible to member of friend
};

Notes

Friendship is not transitive (a friend of your friend is not your friend)

Friendship is not inherited (your friend's children are not your friends)

Storage class specifiers are not allowed in friend function declarations. A function that is defined in the friend declaration has external linkage, a function that was previously defined, keeps the linkage it was defined with.

Access specifiers have no effect on the meaning of friend declarations (they can appear in private: or in public: sections, with no difference)

A friend class declaration cannot define a new class (friend class X {}; is an error)

When a local class declares an unqualified function or class as a friend, only functions and classes in the innermost non-class scope are looked up, not the global functions:

class F {};
int f();
int main()
{
    extern int g();
    class Local { // Local class in the main() function
        friend int f(); // Error, no such function declared in main()
        friend int g(); // OK, there is a declaration for g in main()
        friend class F; // friends a local F (defined later)
        friend class ::F; // friends the global F
    };
    class F {}; // local F
}

A name first declared in a friend declaration within class or class template X becomes a member of the innermost enclosing namespace of X, but is not accessible for lookup (except argument-dependent lookup that considers X) unless a matching declaration at the namespace scope is provided - see namespaces for details.

Template friends

Both function template and class template declarations may appear with the friend specifier in any non-local class or class template (although only function templates may be defined within the class or class template that is granting friendship). In this case, every specialization of the template becomes a friend, whether it is implicitly instantiated, partially specialized, or explicitly specialized.

class A {
    template<typename T>
    friend class B; // every B<T> is a friend of A
 
    template<typename T>
    friend void f(T) {} // every f<T> is a friend of A
};

Friend declarations cannot refer to partial specializations, but can refer to full specializations:

template<class T> class A {}; // primary
template<class T> class A<T*> {}; // partial
template<> class A<int> {}; // full
class X {
    template<class T> friend class A<T*>; // error!
    friend class A<int>; // OK
};

When a friend declaration refers to a full specialization of a function template, the keyword inline and default arguments cannot be used.

template<class T> void f(int);
template<> void f<int>(int);
 
class X {
    friend void f<int>(int x = 1); // error: default args not allowed
};

If a member of a class template A is declared to be a friend of a non-template class B, the corresponding member of every specialization of A becomes a friend of B. If A is explicitly specialized, as long as there is a member of the same name, same kind (type, function, class template, function template), same parameters/signature, it will be a friend of B.

template<typename T> // primary template
struct A
{
    struct C {};
    void f();
    struct D {
        void g();
    };
};
 
template<> // full specialization
struct A<int>
{
    struct C {};
    int f();
    struct D {
        void g();
    };
};
 
class B // non-template class
{
    template<class T>
    friend struct A<T>::C; // A<int>::C is a friend, as well as all A<T>::C
 
    template<class T>
    friend void A<T>::f(); // A<int>::f() is not a friend, because the
                           // signatures do not match, but A<char>::f() is
 
    template<class T>
    friend void A<T>::D::g(); // A<int>::D::g() is not a friend: it is not a member
                              // of A, and A<int>::D is not a specialization of A<T>::D
};

Template friend operators

A common use case for template friends is declaration of a non-member operator overload that acts on a class template, e.g. operator<<(std::ostream&, const Foo<T>&) for some user-defined Foo<T>

Such operator can be defined in the class body, which has the effect of generating a separate non-template operator<< for each T and makes that non-template operator<< a friend of its Foo<T>

#include <iostream>
 
template<typename T>
class Foo {
 public:
    Foo(const T& val) : data(val) {}
 private:
    T data;
 
    // generates a non-template operator<< for this T
    friend std::ostream& operator<<(std::ostream& os, const Foo& obj)
    {
        return os << obj.data;
    }
};
 
int main()
{
    Foo<double> obj(1.23);
    std::cout << obj << '\n';
}

1.23

or the function template has to be declared as a template before the class body, in which case the friend declaration within Foo<T> can refer to the full specialization of operator<< for its T:

#include <iostream>
 
template<typename T>
class Foo; // forward declare to make function declaration possible
 
template<typename T> // declaration
std::ostream& operator<<(std::ostream&, const Foo<T>&);
 
template<typename T>
class Foo {
 public:
    Foo(const T& val) : data(val) {}
 private:
    T data;
 
    // refers to a full specialization for this particular T 
    friend std::ostream& operator<< <> (std::ostream&, const Foo&);
    // note: this relies on template argument deduction in declarations
    // can also specify the template argument with operator<< <T>"
};
 
// definition
template<typename T>
std::ostream& operator<<(std::ostream& os, const Foo<T>& obj)
{
    return os << obj.data;
}
 
int main()
{
    Foo<double> obj(1.23);
    std::cout << obj << '\n';
}

Example

#include <iostream>
#include <sstream>
 
class MyClass {
    int i;
 
    friend std::ostream& operator<<(std::ostream& out, const MyClass& o);
    friend std::istream& operator>>(std::istream& in, MyClass& o);
 public:
    MyClass(int i = 0) : i(i) {}
};
 
std::ostream& operator<<(std::ostream& out, const MyClass& mc)
{
    return out << mc.i;
}
 
std::istream& operator>>(std::istream& in, MyClass& mc)
{
    return in >> mc.i;
}
 
int main()
{
    MyClass mc(7);
    std::cout << mc << '\n';
    std::istringstream("100") >> mc;
    std::cout << mc << '\n';
}

7
100

Defect reports

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.

References

See Also