Query on friend function

Question

/*

How to get out like below using single friend function?

Note: Don't use public getter method.

show value = 100

show value = 1000

*/

#include <iostream>

using namespace std;

class T

{

public:

virtual int show()=0;

virtual ~T()

{}

};

class Test:public T

{

public:

int i;

Test(int _i):i(_i)

{

}

int show()

{

return i;

}

friend void display(T& );

};

class Test2:public T

{

int i;

public:

Test2(int _i):i(_i)

{

}

int show()

{

return i;

}

friend void display(T& );

};

void display(T& t)

{

cout<<"show value = "<<t.i<<endl;

}

int main(void) {

Test t(100);

display(t);

Test2 t1(1000);

display(t1);

return 0;

}

Answer 1

First of all polymorphism doesn't work well with references, use pointers.
The best solution is to use your virtual show from a base pointer and forget the friend function.

Another way is to use CRTP but requires advanced knowledge and it's really complicated.

template <typename D>
Class Base{

friend void display<D>(Base*)
};

class Derived : public Base<Derived>{
int i;
};

template<typename T>
void display (Base* b){
cout<< static_cast<T*>(b)->i;
}

not tested, I suggest you to use your virtual function which is intended for that.