Dynamic Method Dispatch in java

Question

class A{
int a=10;   
public void show(){ 
    System.out.println("Show A: "+a);
    }
}

class B extends A{
public int b=20;    
public void show(){
    System.out.println("Show B: "+b);
    }
}



public class DynamicMethodDispatch {

    public static void main(String[] args) {

    A aObj = new A();       
    aObj.show();    //output - 10

    B bObj = new B();
    bObj.show();   //output - 20

    aObj = bObj;   //assigning the B obj to A..         
    aObj.show();  //output - 20 

    aObj = new B();
    aObj.show();  //output - 20
             System.out.println(bObj.b);  //output - 20 
    //System.out.println(aObj.b); //It is giving error



     }
}

In the above program i'm getting Error wen i try invoking aObj.b.
1.why i'm not able to acess that variable through the aObj though it is refering to class B??
2. why i'm able to acess the method show()?

aioobe · Accepted Answer · 2012-05-15 07:06:40Z

5

You have to distinguish between the static type of aObj and the runtime type of aObj.

Code such as

A aObj = new B();

results in an aObj variable with static type A and runtime type B.

The compiler will only bother too look at the static type when deciding what to allow or not.

To your questions:

1.why i'm not able to acess that variable through the aObj though it is refering to class B??

Because there is (in general) no way for the compiler to know that aObj will refer to a B object at runtime, only that it will refer to some form of A object. Since .b is not available on all A objects, so the compiler will think "better safe than sorry" and disallow it.

2.why i'm able to acess the method show()?

Because this method is available in all A objects (if it's not declared in the subclass, it is still inherited from A).

edited May 15, 2012 at 7:06

answered May 15, 2012 at 7:00

aioobe

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1 Comment

anon Over a year ago

The first 4 lines of your answer were invaluable for me. Thanks.

ftr · Accepted Answer · 2012-05-15 07:02:13Z

2

aObj is a local variable of type A. A has no member called b, that's only in your subclass B. If you want to use b, you need to declare a variable of type B, but of course you can only assign it instances of B (or subclasses if there are any).

A declares a method show(), but you override the implementation in your subclass B.

edited May 15, 2012 at 7:02

answered May 15, 2012 at 6:54

ftr

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4 Comments

Jeevi Over a year ago

But How it is accessing the show() of B? aObj = bObj; aObj.show(); //output - 20 which is the value of b

ftr Over a year ago

A declares a method show(), but you override the implementation in your subclass B. You probably should look into the concepts of OO/inheritance a bit further.

Jeevi Over a year ago

Ya.. i got it..Yes., i'm really poor in opp principles.. Where to read the OOP principle? recommend some good sites.. Thanks

Jeevi Over a year ago

i didnt do it dude... Belive me.. I don have sufficient points to do tat also.

HashimR · Accepted Answer · 2012-05-15 07:05:36Z

1

This behavior is known as virtual method invocation, and it is an important aspect of polymorphism in Java. You should have a look at this tutorial.

answered May 15, 2012 at 7:05

HashimR

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Comments

Kshitij · Accepted Answer · 2012-05-15 07:05:53Z

class A{ // class A has variable a and method show();
int a=10;   
public void show(){ 
    System.out.println("Show A: "+a);
    }
}

class B extends A{ //class B inherits variables and methods of A.
   // so class B has variable a, b and show(). Also, show is overridden by class B.
public int b=20;    
public void show(){
    System.out.println("Show B: "+b);
    }
}

since A doesn't have variable b inside it, even when u are passing B to A, you still have A object which does not have variable b inside it. So, trying to access b will give you compile time error.
in case of show(), A and B both have this method, so what you are doing here is actually overriding it at runtime. This is nothing but Polymorphism. So since A already has method show(), and it is overridden later by B,

A a = new B(); a.show();

this will run the show() method of B at runtime.

Sridhar · Accepted Answer · 2012-05-15 07:10:39Z

0

Methods and fields have different polymorphic behaviour.

The method that will be called is the method of the run time type of the instance

aObj=new B(); //new B()

The field that will be called is the field of the type of reference that you declared

A aObj = new A(); // A aObj

The following would work even is there was no show() method in A.

aObj = new B();
aObj.show(); //calls B's show()

edited May 15, 2012 at 7:10

answered May 15, 2012 at 7:03

Sridhar

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1 Comment

Akki Over a year ago

I checked out the example in JAVA, if i try A aobj=new B() and then aobj.show(), then compiler gives error asking me to define method show() in A

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