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<< on different objects (e.g. output to file etc), but wants to make a cout that isn't cout. — Tas
– Tas, Commented Jul 25, 2016 at 3:22

Indiana Kernick · Accepted Answer · 2016-07-25 03:33:35Z

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

best way is to make operator << return reference, like this: MyClass &operator << (...); so you would be able to write several << in one line: myObject << 42 << 37 << 14;.

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