CODEWITHSUNDEEP
c++iostreamostreaminsertion
Daniel
Daniel

Reputation: 2592

C++ insertion operator for class method

In C++ is there a way to use the insertion operator for a class method?

This operator<< overload is working:

class Complex {
public:
  //Normal overload:
  friend std::ostream& operator<<(std::ostream &out, const Complex &o) {
    out << "test overload";
    return out;
  }

  Complex() {};
  ~Complex() {};
};

I can do this:

int main()
{
  Complex* o = new Complex();

  std::cout << "This is test: " << *o << "." << std::endl; // => This is test: test overload.
}

I know about stream manipulators, like this:

std::ostream& welcome(std::ostream& out)
{
    int data = 1;
    out << "WELCOME " << data << "\r\n";
    return out;
}

int main()
{
  std::cout << "Hello " << welcome; // => "Hello WELCOME 1\r\n"
}

How can I put the welcome method into the Complex class and then how shall I call it from cout (please note that welcome method must access some class member variables)?

My trial:

class Complex {
public:
  //Normal overload:
  friend std::ostream& operator<<(std::ostream &out, const Complex &o) {
    out << "test overload";
    return out;
  }

  std::ostream& welcome(std::ostream& out) {
    out << "WELCOME " << data << "\r\n";
    return out;
  }

  Complex() { data = 1; };
  ~Complex() {};
private:
  int data;
};

int main()
{
  Complex* o = new Complex();

  std::cout << "This is test2: " << o->welcome << std::endl; // compile error
}

Upvotes: 0

Views: 226

Answers (1)

463035818_is_not_an_ai
463035818_is_not_an_ai

Reputation: 122595

One easy way to pick a different << overload is to use a different type.

#include <iostream>

class Complex {
public:
  //Normal overload:
  friend std::ostream& operator<<(std::ostream &out, const Complex &o) {
    out << "test overload";
    return out;
  }

  struct extra_info {
      const Complex& parent;
      extra_info(const Complex& p) : parent(p) {}
      friend std::ostream& operator<<(std::ostream& out, const extra_info& ei){
        int i = 1;
        out << "extrainfo " << i;
        return out;
      }
  };
  extra_info extrainfo() {
      return {*this};
  }

  Complex() {};
  ~Complex() {};
};


int main() {
    Complex c;
    std::cout << c << "\n";
    std::cout << c.extrainfo();
}

Output:

test overload
extrainfo 1

I suppose in your real code you are using members. Hence the helper type must hold a reference to the Complex instance.

Upvotes: 2

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