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Friday, August 03, 2007

Parameterized Constructors of Derived Classes

In the previous article Constructors and Destructors of Derived Classes, we’re discussing about the calling conventions of Constructors and Destructor functions of derived classes whose base class also had them. There was one thing special about those constructors; none of them were taking arguments.

If only the derived’s constructor takes parameters then also its O. K. but what if both the base and derived class contains parameterized constructors (as is obvious from the code below).


  // this code contains ERRORS

  // base class
  class base
  {
    int a;

  public:
    base(int n)
    {
      //...
    }
  };

  // derived class
  class derived:public base
  {
    int b;

  public:
    derived(int m)
    {
      //...
    }
  };

  //main
  void main()
  {
    base b(10); //ok

    derived d(10); // ERROR!!
      // base's constructor
      // will also be called and
      // it needs parameters too !!
  }

How’d you pass arguments to the base class constructor?

To answer this, we need to introduce Expanded from of Constructor Declaration (of the derived class).


  derived-constructor (arg): base1 (arg),
                             base2 (arg),
                             ...
                             baseN (arg)
  {
    ...
    ...
  }

Here base1, base2 etc. are constructor functions of base classes, which the derived class inherits.

Let’s have a look at an example program to understand this declaration:


  // this code is ok

  // base class
  class base
  {
    int a;

  public:
    base(int n){a=n;}
  };

  // derived class
  class derived:public base
  {
    int b;

  public:
    // since this class is derived
    // from only one class 'base'
    // therefore only one constructor
    // is listed
    derived(int m1,int m2):base(m2)
    {b=m1;}
  };

  // main
  void main()
  {
    base b(10); //ok

    derived d(10,100); //ok
  }

One thing to note here is, the argument which is to be passed to the base’s constructor should be declared and accepted from the derived class constructor, as in the line

derived(int m1,int m2):base(m2)
{b=m1;} 

Here derived’s constructor is only using the parameter m1 but is accepting another parameter m2 which is to be passed along to the constructor of the base class.

Let us have a look at another example program which will clear all the confusions. Keep reading the comments!


  #include<iostream.h>

  // base class (1)
  class base1
  {
  protected:
    int a;

  public:
    base1(int p1){a=p1;}
  };

  //base class (2)
  class base2
  {
  protected:
    int b;

  public:
    base2(int p2){b=p2;}
  };

  // derived class (1)
  class derived1:public base1
  {
  protected:
    int c;

  public:
    derived1(int p3,int p4):base1(p4)
    {c=p3;}

    void show(){cout<<a<<"\n"<<c;}
  };

  //derived class (2)
  class derived2:public base1, public base2
  {
  public:
    // this class inherits two base classes
    // therefore we are using two base
    // class constructors in the declaration
    derived2(int p5,int p6):base1(p5),
                base2(p6)
    {
      // empty

      // used only to call the base's
      // constructors with the passed
      // arguments

      // derived class constructor may not
      // use any arguments passed to it
    }

    void show(){cout<<a<<"\n"<<b;;}
  };

  // main
  void main()
  {
    derived1 d1(10,20);
    derived2 d2(100,200);

    d1.show();
    cout<<endl;
    d2.show();
  }

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3 comments:

  1. Anonymous6:16 PM

    oh what a nice study materail i like it

    ReplyDelete
  2. Anonymous8:12 PM

    oh it is really helpful to us

    ReplyDelete
  3. Thanx lots

    -Vijay wagh
    from jalna

    ReplyDelete

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