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Monday, August 13, 2007

Introduction to Operator Overloading in C++ II

From the previous article Introduction to Operator Overloading in C++, we know that we can overload operators by defining member functions having special names. The general form, as you know, for overloading operators is:

ret-type operator#(arg-list);

Although, you’re free to use any return type as ret-type but commonly it’ll be the name of the class itself; in this article we’ll be closely examining why is it so and what would happen if we use other return types.

You may not expect further theories in this article as all the further discussion is in the program as comments.

I request you to read the program thoroughly so you may understand what we’re discussing.


  // Example program to illustrate
  // operator overloading with diff rent
  // return-types used in the overload
  // function 
  #include <iostream.h>

  // class
  class myclass
  {
    int a;

  public:
    // default constructor
    myclass(){}

    // constructor
    myclass(int x){a=x;}

    // overloaded operator
    // with return type 'myclass'
    myclass operator +(myclass);

    void show(){cout<<a<<endl;}
  };

  myclass myclass::operator +(myclass ob)
  {
    myclass t;

    t.a=a + ob.a;

    return t;
  }

  // other class
  class myclass2
  {
    int a;

  public:
    // default constructor
    myclass2(){}

    // constructor
    myclass2(int x){a=x;}

    // here, return type is
    // 'int'
    int operator +(myclass2);

    void show(){cout<<a<<endl;}
  };

  int myclass2::operator +(myclass2 ob)
  {
    int t;

    t=a + ob.a;

    return t;
  }

  void main()
  {
    myclass ob1(10);
    myclass ob2(20);

    myclass2 ob3(100);
    myclass2 ob4(200);

    int a;

    // this is legal because
    // the class to which ob1 and ob2
    // belong, upon '+' operation return
    // data of type 'myclass' 
    ob1=ob1+ob2;

    // here the result of the '+'
    // operation return data of type
    // 'int' which is assigned to the
    // variable 'a' (of data type int)
    a=ob3+ob4;

    ob1.show();
    cout<<a;
  }

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3 comments:

  1. Anonymous4:42 PM

    It will be grt if u explain with the output.
    -Manasi

    ReplyDelete
  2. Hi Manasi,

    I guess the program code has enough comments. I have purposefully not discussed the output, as an exercise for the readers.

    ReplyDelete
  3. Anonymous11:51 AM

    Here the out put is 30 only or 30,300

    ReplyDelete

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