How and why inheritance of methods differ from inheritance of variables?

Question

while preparing myself for OCJP exam i stumbled upon one question i am unable to wrap my mind around. Here's (a bit modified by me) code from a question:

class Foo {
    public int a = 3;
    public void addFive() {
        a += 5;
        System.out.print("f ");
    }
}

class Bar extends Foo {
    public int a = 8;
    public void addFive() {
        a += 5;
        System.out.print("b ");
    }
}

public class TestInheritance {
    public static void main(String [] args) {
        // part 1
        Foo f = new Bar();
        f.addFive();
        System.out.println(f.a);

        // part 2
        Bar b = new Bar();
        b.addFive();
        System.out.println(b.a);
    }
}

and the output is:

b 3
b 13

Part 2 i can understand. Nothing surprising here. However part 1 does not let me sleep at night. I understand why Bar.addFive was run but why in part 1 f.a prints Foo.a when I used new Bar() to instantiate an object? It looks like inheritance works quite different for methods than for variables. What am i missing here to understand that concept? What am I failing in?

possible duplicate of Hiding instance variables of a class

Brett Okken
– Brett Okken

2014-07-19 15:52:19 +00:00
Commented Jul 19, 2014 at 15:52 — Brett Okken
– Brett Okken, Commented Jul 19, 2014 at 15:52

Sergey Kalinichenko · Accepted Answer · 2014-07-19 15:54:31Z

3

It looks like inheritance works quite different for methods then for variables.

More precisely, inheritance does not make variables polymorphic. When you declare an identical method, a method in the derived class overrides the method in the base. When you declare an identical variable, the variable in the base class is hidden by the variable in the derived class.

In your example, Bar has two variables a, but only one method addFive(). Essentially, each Bar object holds two integers - Foo.a and Bar.a. These two are separate variables. However, the addFive method is the one defined in Bar, because it overrides (replaces) addFive from Foo.

Note that Java lets Bar access Foo.addFive by calling super.addFive(), but users of Bar lack this possibility.

answered Jul 19, 2014 at 15:54

Sergey Kalinichenko

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Comments

Kyle Emmanuel · Accepted Answer · 2014-07-19 15:56:12Z

2

Variables are not polymorphic. only methods are.

However you can access them from the parent class if they are declared protected or public or through setters and getters

answered Jul 19, 2014 at 15:56

Kyle Emmanuel

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Suresh Atta · Accepted Answer · 2014-07-19 15:51:28Z

0

 Foo f = new Bar();

That line alone means Access the implementations in Bar and members from Foo :)

So you'll get a from Foo and the method implementations from Bar

answered Jul 19, 2014 at 15:51

Suresh Atta

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Ganesh Chithambalam A.S. · Accepted Answer · 2014-07-19 16:02:11Z

0

In order to understand the part 1 you need to understand the difference between static and dynamic binding.

Static binding occurs during compile time and private, final and static methods and variables uses static binding. Static binding uses Type(Class) information for binding.

Based on the above definition when the line Foo f = new Bar(); get compiled, the variable a is initialized 3.

Methods are resolved using Dynamic binding and Dynamic binding uses Object to resolve binding. Since the object is of the Class Bar() the addFive() method of the Bar class is executed. Hence the output b3

answered Jul 19, 2014 at 16:02

Ganesh Chithambalam A.S.

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