Is it possible to use the class object as a value? [duplicate] - c++

Question

This question already has answers here:
How can I use cout << myclass
(5 answers)
Closed 6 years ago.
I don't know the term in programing of what I'm asking (this is hobby for me) and I'm trying to new things here. See my working escenario:
#include <iostream>
class Foo {
int _x;
public:
Foo () : _x(0) {}
Foo (int x) : _x(x) {}
int get () const { return _x; }
};
int main () {
Foo f1;
Foo f2(10);
std::cout << "Value: " << f1.get () << std::endl; // 0
std::cout << "Value: " << f2.get () << std::endl; // 10
return 0;
}
Is it possible to use either f1 or f2 like this:
std::cout << "Value: " << f2 << std::endl; // shows 10
updated with correct code:
#include <iostream>
class Foo {
int _x;
public:
Foo () : _x(0) {}
Foo (int x) : _x(x) {}
int get () const { return _x; }
friend std::ostream &operator<<(std::ostream &os, const Foo& f) {
return os << f.get ();
}
};
int main () {
Foo f1;
Foo f2(10);
std::cout << "Value: " << f1.get () << '\n'; // 0
std::cout << "Value: " << f2.get () << '\n'; // 10
std::cout << "Value: " << f1 << '\n'; // 0
return 0;
}

Answer 1

Yes, this is overloading stream insertion operator.
#include <iostream>
class Foo {
int _x;
public:
Foo () : _x(0) {}
Foo (int x) : _x(x) {}
int get () const { return _x; }
friend std::ostream& operator<< ( std::ostream& stream, const Foo& foo );
};
std::ostream& operator<< ( std::ostream& stream, const Foo& foo ) {
stream << foo._x;
return stream;
}
int main () {
Foo f1;
Foo f2(10);
std::cout << "Value: " << f1 << std::endl; // 0
std::cout << "Value: " << f2 << std::endl; // 10
return 0;
}