Question

So I have class A and class B, where class B extends class A. I must overload the << and >> in both classes. I was hoping that in the function definition of the operators for class B, I could call the overloaded operators from class A, but I’m having trouble doing so.

#include <iostream>
#include <string>
using namespace std;

class A {
friend ostream& operator<<(ostream& out, A a);
protected:
    int i;
    string st;
public:
    A(){
        i=50;
        st = "boop1";
    }
};

ostream& operator<<(ostream &out, A a) {
out << a.i << a.st;
return out;
}

class B : public A {
friend ostream& operator<<(ostream& out, B b);
private:
    int r;
public:
    B() : A() {
        r=12;
    }
};

ostream& operator<<(ostream &out, B b) {
out = A::operator<<(out, b);    //operator<< is not a member of A
out << "boop2" << b.r;
return out;
}

int main () {
B b;
cout << b;
}

I attempt to call A’s version of operator<< in B’s version of operator<<, but of course it doesn’t actually belong to A, so it cannot compile. How should I be achieving this?

also, note that in reality A and B have their own header and body files.

Answer 1

You can make your B object look like an A object:

std::ostream& operator<< (std::ostream& out, B const& b) {
    out << static_cast<A const&>(b);
    out << "boop2" << b.r;
    return out;
}

Note that you almost certainly don’t want to pass the object to be printed by value. I have changed the signature to use a const& instead: This indicates that the object won’t get changed and it won’t get copied.