friend-function

This is a follow-up on my answer to this question . The original answer I posted does not have any namespaces, and it solves the problem. However, the OP subsequently made an edit claiming that the solution does not work if the class is inside a namespace. My initial reaction was that you can make it work by simply having using N::f; either at global scope or inside main() (or any other function). While that is certainly the case, the OP (justifiably) commented that this is not ideal, and I agree. Nevertheless, I still thought that calling N::f without having using N::f; should work just fine,

Minimal example: class A { friend void swap(A& first, A& second) {} void swap(A& other) {} void call_swap(A& other) { swap(*this, other); } }; int main() { return 0; } g++ 4.7 says: friend.cpp: In member function ‘void A::call_swap(A&)’: friend.cpp:7:20: error: no matching function for call to ‘A::swap(A&, A&)’ friend.cpp:7:20: note: candidate is: friend.cpp:4:7: note: void A::swap(A&) friend.cpp:4:7: note: candidate expects 1 argument, 2 provided Outcomment line 4: // void swap(A& other) {} ...and it works fine. Why, and how to fix this, if I want to keep both variants of my swap function? I

I have a doubt related to friend functions in C++. Friend function is not a member function of the claas and can be invoked directly from the main. So, what difference does it make if we keep the friend function within the private or the public part of the class . I have generally noticed that the friend functions are always in the public part. In what scenario we should keep the friend function within private . The compiler does not pay any attention to whether a friend function is in the private or public (or protected) section of a class. Most people put it in the public section, but it'll

I'm having a situation similar to the one described in Specify a class member function as a friend of another class? . However in my case, class B needs to know class A since it's using it, so the solution given in that thread is not working for me. I tried to give also a forward declaration to the function itself but it didn't work as well. It seems that each class need the full definition of the other... Is there any easy way to solve it? I prefer a solution which doesn't involve new classes which wrap one of the old classes. code example: //A.h class B; //not helping void B::fB(); //not

#include <iostream> class B; class A{ int a; public: friend void B::frndA(); }; class B{ int b; public: void frndA(); }; void B::frndA(){ A obj; std::cout << "A.a = " << obj.a << std::endl; } int main() { return 0; } When trying to compile this code, some errors occurred. E.g. invalid use of incomplete type What are the problems in this code? Place the whole of the class B ... declaration before class A . You haven't declared B::frndA(); yet. #include <iostream> using namespace std; class B{ int b; public: void frndA(); }; class A{ int a; public: friend void B::frndA(); }; void B::frndA(){ A

Today i have a doubt regarding friend function. Can two classes have same friend function? Say example friend void f1(); declared in class A and class B. Is this possible? If so, can a function f1() can access the members of two classes? An example will explain this best: class B; //defined later void add(A,B); class A{ private: int a; public: A(){ a = 100; } friend void add(A,B); }; class B{ private: int b; public: B(){ b = 100; } friend void add(A,B); }; void add (A Aobj, B Bobj){ cout << (Aobj.a + Bobj.b); } main(){ A A1; B B1; add(A1,B1); return 0; } Hope this helps! There is no