Question

I would like to create a c++ type that mimic the build-in type exactly. Below is an example of an “integer” type that “boxed in” an “int” type. Problem I have is, I want to show the value of “integer” using only the stand alone “integer” object d, such that cout << d will show the value, not cout << d.show ( ); how would I do that ?

#include <iostream>

class integer {

      public:
             integer (int x)  { i = x; };
             integer ( ) { };   // default constructor

      integer operator+ (integer& c ){
              return integer(i + c.i);
          }


      int show ( ) { return i; }


      private:
              int i;
};



int main ( ) {

    integer i = 5;
    integer c (10);
    integer d;

    d = i + c;

    std::cout << d.show() << std::endl;

    std::cin.get();
    return 0;   
}

Answer 1

You can overload operator << to do that:

ostream& operator <<(ostream& stream, const integer& myInteger)
{
    return stream << myInteger.show();
}

and make show const.