No, it’s deliberately impossibile. Is an issue by encapsulation. Suppose to have a class “PswClass” that manage any password, that is cascade friend with other class: if I inherit from PswClass: class Myclass : public PswClass { ……. } In this way I can, maybe, have access to field that it would be private.
Is it not supposed for a friend function to be explicitly defined outside of a class ? Friend functions can be defined (given a function body) inside class declarations. These functions are inline functions, and like member inline functions they behave as though they were defined immediately after all class members have been seen but … Read more
This is one of those frequently asked questions that have different approaches that are similar but not really the same. The three approaches differ in who you are declaring to be a friend of your function –and then on how you implement it. The extrovert Declare all instantiations of the template as friends. This is … Read more
When you declare a friend function with an unqualified id in a class it names a function in the nearest enclosing namespace scope. If that function hasn’t previously been declared then the friend declaration doesn’t make that function visible in that scope for normal lookup. It does make the declared function visible to argument-dependent lookup. … Read more
I agree with the comments that say the friend keyword can improve encapsulation if used wisely. I’d just add that the most common (legitimate!) use for friend classes may be testing. You may want a tester class to have a greater degree of access than other client classes would have. A tester class could have … Read more
class A{ public: friend void fun(A a){std::cout << “Im here” << std::endl;} friend void fun2(){ std::cout << “Im here2″ << std::endl; } friend void fun3(); }; Although your definition of fun2 does define a “global” function rather than a member, and makes it a friend of A at the same time, you are still missing … Read more
Quoting N3691 – §11.3/4 [class.friend] A function first declared in a friend declaration has external linkage (3.5). Otherwise, the function retains its previous linkage (7.1.1). So you need to declare the function as static prior to declaring it as a friend. This can be done by adding the following declarations above the definition of A. … Read more
I think you can use Attorney-Client here. In your case example should be like this class FooA { private: void Hello(); void Hello2(); void Hello3(); int m_iData; friend class Client; }; class Client { private: static void Hello(FooA& obj) { obj.Hello(); } static void Hello2(FooA& obj) { obj.Hello2(); } friend class FooB; }; class FooB … Read more
You need a few forward declarations: template <typename T> class Obj; template <typename T> Obj<T> make_obj(T t); template <typename T> class Obj { private: T & t; Obj (T & t) : t(t) { } Obj() = delete; friend Obj make_obj<T>(T t); }; template <typename T> Obj<T> make_obj(T t) { return Obj<T>(t); } live example … Read more