Question

I am trying to overload << operator, so that when I use my class object in std::cout, it prints out the data members which is supposed to get printed.

I know I have to define a friend functions signature inside of a class, and then outside of the class I define two friend functions doing the job, but I want to hand over the class, and don’t want the user to do literally anything and I want the class take care of everything.

How can I achieve such a thing in C++?

Answer 1

C++ is designed to allow non-member functions to do things that would require a member function in other languages. A friend function is very similar to a member function, in that it can be defined within a class scope. Also (and this can be confusing) if you declare a friend function only within a class scope, it can only be accessed by passing an object of that class to it.

struct printme {
    friend std::ostream &operator<< ( std::ostream &os, printme const & )
        { return os << "hello"; }

    friend void named_friend( printme const & ) {}
};

std::cout << printme(); // OK
named_friend( printme() ); // OK
void (*fn_ptr) = named_friend; /* function not found
           it can only be found by association with printme */

Although this subtle rule is odd at first, it reflects the idea that nonmember functions can also be encapsulated by the class.

For more info, read up on argument-dependent lookup (ADL).