Question

I’ve come across functions that, rather than overloading the operator << to use with cout, declare a function that takes an ostream and returns an ostream

Example:

#include <iostream>

class A {
private:
  int number;
public:
  A(int n) : number(n) {}
  ~A() {}
  std::ostream& print(std::ostream& os) const;
  friend std::ostream& operator<<(std::ostream& os, const A a);
};

An example of implementation:

std::ostream& A::print(std::ostream& os) const {
  os << "number " << number;
  return os;
}

std::ostream& operator<<(std::ostream& os, const A a) {
  os << "number " << a.number;
  return os;
}

Now if I run this I can use the different functions in different situations.. E.g.

int main() {
  A a(1);

  std::cout << "Object.";  a.print(std::cout);
  std::cout << "\n\n";
  std::cout << "Object." << a;
  std::cout << "\n\n";

  return 0;
}

Output:

Object.number 1

Object.number 1

There doesn’t seem to be a situation where the print function would be needed since you could only use separately or in the beginning of a “cout chain” but never in the middle or end of it which perhaps makes it useless. Wouldn’t it (if no other use is found) be better off using a function “void print()” instead?

Answer 1

It would make a lot more sense if operator<<() actually looked like

std::ostream& operator<<(std::ostream& os, const A a) {
    return a.print(os);
}

Then operator<<() wouldn’t need to be a friend.