c++ two classes as each others' friends

In this case, Box need access to Cup.func, AND Cup need access to Box.func, both of which are private because I don't want any other class to have access to neither Box.func nor Cup.func.

So they need to declare each other as friend.

Suppose you organise code like Box1.h, Box1.cpp, Cup1.h, Cup1.cpp, it would cause a problem like this: http://www.cnblogs.com/qrlozte/p/4099032.html

The compiler will complain. So the solution is also simple, see Cup2.h, Cup2.cpp, other files remain unchanged. (Of couse, you can change Box1.h Box1.cpp using the same pattern as Cup2.h Cup2.cpp, although it would make your code looks more consistent, but only changing Cup1.h, Cup1.cpp is already enough to satisfy your need.)

Another question, what if Box.doSomething is a function template? If you change main1.cpp to main2.cpp, and Box1.h, Box1.cpp to Box2.h, Box2.cpp, the linker will complain that it cannot find the implementation of Box.doSomething (undefined reference error) (after all, the compiler need all the detail of a template to generate code, if you hide the implementation of Box.doSomething in Box2.cpp, how can you expect the compiler can generate code for Box.doSomething<int>() in main2.cpp when main2.cpp doesn't include Box2.cpp, right?).

So, because function template (and class template) must define (just declare is NOT enough) in the header, you might modify Box2.h and Box2.cpp as shown in Box3.h and Box3.cpp. And you'll find the problem is solved! (Now we got main2.cpp, Box3.h, Box3.cpp, Cup2.h, Cup2.cpp compiled successfully).

Now consider what if Cup.doSomething also need to be a function template? (The same as Box.doSomething in Box3.h). And... yes, you have to include the definition of Cup.doSomething in Cup.h, too!

And you if you that, the compiler will complain Box is an incomplete type thus in Cup.doSomething, "b.func" cannot be resolved. As shown in main3.cpp, Cup3.h, Cup3.cpp.

The reason is because in Cup3.h "class Box;" it indeed declares class Box, but just the class name, no information is provided about the interface of the class at all.

Can you replace "class Box;" with "#include "Box.h"" ? No. Because that'll cause the same problem as Box1.h, Box1.cpp, Cup1.h, Cup1.cpp caused.

The solution ? At least for now, I don't know(If I can seperate Cup.doSomething() into two parts, one part is template and one part is normal function and the template part doesn't have to access memebers of the parameter 'b', then there's a simple solution, just let the template part in the header, and implement the normal function in Cup.cpp). This situation is not made up by me, I really encountered this problem when I was making a little program. Maybe there's something wrong with my design.

main1.cpp

 #include "Box.h"
 #include "Cup.h"

 int main()
 {
     Cup c;
     Box b;
     c.doSomething(b);
     b.doSomething(c);
     return ;
 }

Box1.h

 #ifndef BOX_H
 #define BOX_H

 #include "Cup.h"

 class Box
 {
     friend class Cup;
     public:
         Box();
         ~Box();
         void doSomething(const Cup &c);
     private:
         void func() const;// only visible to Cup
 };

 #endif // BOX_H

Box1.cpp

 #include "Box.h"

 #include <iostream>

 Box::Box()
 {

 }
 Box::~Box()
 {

 }
 void Box::doSomething(const Cup &c)
 {
     c.func();
 }
 void Box::func() const
 {
     using namespace std;
     cout << "Box.func" << endl;
 }

Cup1.h

 #ifndef CUP_H
 #define CUP_H

 #include "Box.h"

 class Cup
 {
     friend class Box;
     public:
         Cup();
         ~Cup();
         void doSomething(const Box &b);
     private:
         void func() const; // only visible to Box
 };

 #endif // CUP_H

Cup1.cpp

 #include "Cup.h"

 #include <iostream>

 Cup::Cup()
 {

 }
 Cup::~Cup()
 {

 }
 void Cup::doSomething(const Box &b)
 {
     b.func();
 }

 void Cup::func() const
 {
     using namespace std;
     cout << "Cup.func" << endl;
 }

Cup2.h

 #ifndef CUP_H
 #define CUP_H

 class Box;

 class Cup
 {
     friend class Box;
     public:
         Cup();
         ~Cup();
         void doSomething(const Box &b);
     private:
         void func() const; // only visible to Box
 };

 #endif // CUP_H

Cup2.cpp

 #include "Cup.h"

 #include "Box.h"

 #include <iostream>

 Cup::Cup()
 {

 }
 Cup::~Cup()
 {

 }
 void Cup::doSomething(const Box &b)
 {
     b.func();
 }

 void Cup::func() const
 {
     using namespace std;
     cout << "Cup.func" << endl;
 }

main2.cpp

 #include "Box.h"
 #include "Cup.h"

 int main()
 {
     Cup c;
     Box b;
     c.doSomething(b);
     b.doSomething<int>(, c);
     return ;
 }

Box2.h

#ifndef BOX_H
#define BOX_H

#include "Cup.h"

class Box
{
    friend class Cup;
    public:
        Box();
        ~Box();
        template <typename T> void doSomething(const T &obj, const Cup &c);
    private:
        void func() const;// only visible to Cup
};

#endif // BOX_H

Box2.cpp

 #include "Box.h"

 #include <iostream>

 Box::Box()
 {

 }
 Box::~Box()
 {

 }
 template <typename T> void doSomething(const T &obj, const Cup &c)
 {
     c.func();
 }
 void Box::func() const
 {
     using namespace std;
     cout << "Box.func" << endl;
 }

Box3.h

 #ifndef BOX_H
 #define BOX_H

 #include "Cup.h"

 class Box
 {
     friend class Cup;
     public:
         Box();
         ~Box();
         template <typename T> void doSomething(const T &obj, const Cup &c);
     private:
         void func() const;// only visible to Cup
 };

 template <typename T> void Box::doSomething(const T &obj, const Cup &c)
 {
     c.func();
 }

 #endif // BOX_H

Box3.cpp

 #include "Box.h"

 #include <iostream>

 Box::Box()
 {

 }
 Box::~Box()
 {

 }

 void Box::func() const
 {
     using namespace std;
     cout << "Box.func" << endl;
 }

main3.cpp

 #include "Box.h"
 #include "Cup.h"

 int main()
 {
     Cup c;
     Box b;
     c.doSomething<int>(, b);
     b.doSomething<int>(, c);
     return ;
 }

Cup3.h

 #ifndef CUP_H
 #define CUP_H

 class Box;

 class Cup
 {
     friend class Box;
     public:
         Cup();
         ~Cup();
         template <typename T> void doSomething(const T &obj, const Box &b);
     private:
         void func() const; // only visible to Box
 };

 template <typename T> void doSomething(const T &obj, const Box &b)
 {
     b.func();
 }

 #endif // CUP_H

Cup3.cpp

 #include "Cup.h"

 #include "Box.h"

 #include <iostream>

 Cup::Cup()
 {

 }
 Cup::~Cup()
 {

 }

 void Cup::func() const
 {
     using namespace std;
     cout << "Cup.func" << endl;
 }