How to your own C++ cout << like object

Question

I was just asking if you can create your own cout<< like object in C++. most people confuse my question with operator overloading <<. But no, i dont want to implement my own << operator so that when users print my object i can control what they get. But basically i want to implement like this:

something << some_given << some_end;

Not sure if that is possible , but the iostream standard library created the cout , so my mind says "Why not?". So i asked stackoverflow. Help would be appreciated! :)

I think op wants to make a `cout` object, but not `cout`. I think op understands about how to use `operator< — Tas, Jul 25 '16 at 03:22
@Dsafds see http://stackoverflow.com/questions/13703823/a-custom-ostream — kfsone, Jul 25 '16 at 03:26

Indiana Kernick · Accepted Answer · 2016-07-25T03:33:35.770

3

I'm not sure if I interpretted your question correctly but I think you want a class with an overloaded operator<< so that's what i have here

class MyClass {
  public:
    MyClass() = default;
    MyClass& operator<<(int input) {
      //do something with input
      return *this;
    }
}

You would use it like this;

MyClass myObject;
myObject << 42;
//the function would have been called

edited Jul 25 '16 at 03:33

answered Jul 25 '16 at 03:25

Indiana Kernick

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Thanks thats what i wanted! – amanuel2 Jul 25 '16 at 03:28
2

best way is to make `operator < – Andrei R. Jul 25 '16 at 03:31

How to your own C++ cout << like object

1 Answers1