Printing derived object using operator<< overload not working - c++

Question

Why can I not print the derived object using this approach? How can I fix this so that the derived printout prints "derived size=8". I ideally want to keep the printing code out of the class code.
#include <iostream>
class base
{
public:
base(int size) : size_(size) {}
virtual int get_size() const { return size_; }
private:
int size_;
};
std::ostream& operator<<(std::ostream& os, const base& obj) {
os << "base size=" << obj.get_size() << std::endl;
return os;
}
class derived1 : public base
{
public:
derived1(int size) : base(size) {}
virtual int get_size() const {
return 2 * base::get_size();
}
};
std::ostream& operator<<(std::ostream& os, const derived1& obj) {
os << "derived1 size=" << obj.get_size() << std::endl;
return os;
}
int main(int argc, char* argv[]) {
base* b1 = new base(3);
std::cout << "b1 size is: " << b1->get_size() << std::endl;
std::cout << *b1 << std::endl;
base* b2 = new derived1(4);
std::cout << "b2 size is: " << b2->get_size() << std::endl;
std::cout << *b2 << std::endl;
delete b1;
delete b2;
return 0;
}
Output:
b1 size is: 3
base size=3
b2 size is: 8
base size=8
UPDATE:
I changed as follows as per ghostsofstandardspast which works:
#include <iostream>
class base
{
public:
base(int size) : size_(size) {}
virtual int get_size() const { return size_; }
friend std::ostream& operator<<(std::ostream &os, const base& obj);
private:
virtual std::ostream &print(std::ostream &os) const {
return os << "base size=" << get_size() << std::endl;
}
int size_;
};
std::ostream& operator<<(std::ostream& os, const base& obj) {
obj.print(os);
return os;
}
class derived1 : public base
{
public:
derived1(int size) : base(size) {}
virtual int get_size() const {
return 2 * base::get_size();
}
friend std::ostream& operator<<(std::ostream &os, const derived1& obj);
private:
virtual std::ostream &print(std::ostream &os) const {
return os << "derived1 size=" << get_size() << std::endl;
}
};
int main(int argc, char* argv[]) {
base* b1 = new base(3);
std::cout << "b1 size is: " << b1->get_size() << std::endl;
std::cout << *b1 << std::endl;
base* b2 = new derived1(4);
std::cout << "b2 size is: " << b2->get_size() << std::endl;
std::cout << *b2 << std::endl;
return 0;
}

Answer 1

operator<< is not a virtual member function, so it can't be used polymorphically like other member functions can. Instead, you should delegate to a virtual member.
Change both of your operator<< overloads to a virtual member function called print or something similar. Then, overload operator<< to delegate to that:
std::ostream &operator<<(std::ostream &os, const Base &obj) {
obj.print(os); //or whatever name you choose
return os;
}
If you want to make the member function private, make operator<< a friend.
Finally, note that printing a pointer to an object of your class will only print the address. To invoke operator<<, you need to print the object itself:
Base *b = /*whatever*/;
std::cout << *b;
This technique for a polymorphic operator<< is demonstrated in this cppquiz question. I highly recommend going through those questions.

Answer 2

cout << b2 doesn't invoke your operator <<, because b2 is a pointer not a reference. Try cout << *b2 instead.

Answer 3

It doesn't work because in a call
std::cout << b1 << std::endl;
you are printing an address, since b1 is a pointer ( i.e. 0x8e02008).
You should dereference it instead
std::cout << *b1 << std::endl;
In this case both for *b1 and *b2 objects the std::ostream& operator<<(std::ostream& os, const base& obj); will be calles and virtual call to get_size() will be executed.
ideone