Java code explanation

Question

class Base 
{
    int i = 99;
    public void amethod()
    {
        System.out.println("Base.amethod()");
    }

    Base()
    {
        amethod();
    }
}
public class Derived extends Base
{
    int i = -1;

    public static void main(String argv[])
    {
        Base b = new Derived();
        System.out.println(b.i);
        b.amethod();
    }

    public void amethod()
    {
        System.out.println("Derived.amethod()");
    }
}

Why does this code print b.i = 99 and not b.i = -1? Thank you.

Kenny Wyland · Accepted Answer · 2011-03-13 18:23:28Z

4

It's because you are referencing a member variable of the object instead of a method. Since it is declared as a Base object, it will use Base.i because member variables do not benefit from polymorphism that is afforded to methods. If you added a getI() method to both classes and called that instead of just b.i, it would work as you expect.

public int getI() {
    return this.i;
}


    Base b = new Derived();
    System.out.println(b.getI());

answered Mar 13, 2011 at 18:23

Kenny Wyland

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2 Comments

Adrian Over a year ago

and again, why Base() constructor is calling amethod() in Derived? if you can explain

Kenny Wyland Over a year ago

Because when you call a method on an object, it calls the method on the real type of object. So even though you are in the Base constructor, when you call amethod() it begins looking for an amethod() implementation at the child class. If it doesn't find one there, then it'll move up the inheritance hierarchy, but it always tries the child first.

SLaks · Accepted Answer · 2011-03-13 18:19:43Z

2

You have two different fields named i; one in Base and one in Derived.

Since b is declared as a Base, b.i returns the i field from the Base class.

You probably want to set the i field in the Derived constructor instead of creating a new field.

answered Mar 13, 2011 at 18:19

SLaks

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Comments

John Vint · Accepted Answer · 2011-03-13 18:35:00Z

2

Because in this instance b.i is saying ((Base).i) meaning it will reference the Base's i variable and not the Derived i. If instead you did b.getI() and each Base and Dervid implemented their own getI and returned their own i it would return -1.

To explain why Base is calling Derived's amethod. Java method invocation is dynamic and determined at runtime. In your case Base b = new Derived() b here is of type Derived. The Java runtime will realize b's type and invoke the amethod created closest to Derived.

So if you have

class Base{
  int i =99;
  public int getI(){
    return i;
  } 
}
class Derived extends Base{
  int i =-1;
  public int getI(){
    return i;
  }
}

Derived's getI() method will be called (and return -1) even if its defined as Base b = new Derived();

edited Mar 13, 2011 at 18:35

answered Mar 13, 2011 at 18:20

John Vint

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Comments

Mike Samuel · Accepted Answer · 2011-03-13 18:20:02Z

1

Methods can be overridden, but fields mask, so your subclass defines a new field named i. The superclass methods refer to the old field i defined in the superclass.

answered Mar 13, 2011 at 18:20

Mike Samuel

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