Inheritance and public/private methods

Question

One.java

public class One {

    private void run() {
        System.out.println("one");
    }

    public void start() {
        this.run();
    }
}

Two.java

public class Two extends One {
    public void run() {
        System.out.println("two");
    }
}

Main.java

public class Main {
    public static void main(String[] args) {
        Two t = new Two();
        t.start();
    }
}

Output: one

But if I declare the run method public in the One class I get "two" as output.

It is pretty unpredictable, how does this work?

When `One`s `run` method is private, `Two` can't access it. When it's `public` the version in `Two` overrides it. — JonK, Oct 16 '15 at 13:25
@JonK that's how I read your comment. But you've got a lot of up votes so maybe that's just me :) — ControlAltDel, Oct 16 '15 at 13:31

score 6 · Accepted Answer · answered Oct 16 '15 at 13:27

6

The run method in Two will not be overriding the private method in One, because it is private and inaccessible to any class outside One.

Once the method is marked as public, the inheritance system takes place, and run is invoked from start in One (as the method is not overridden), but then resolved dynamically (at runtime) as Two's run.

My advice, use the @Override annotation to make sure at compile-time that a method is properly overridden.

Also, you seem to be playing with start and run methods: remember you need to extend Thread or implement Runnable for the JVM to recognize your classes as Runnables.

answered Oct 16 '15 at 13:27

Mena

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but why in private case `this.run()` give output One's method rather than Two's run because `this` in the `Two` object not `One` – codegasmer Oct 16 '15 at 13:43
2

@codegasmer http://stackoverflow.com/questions/5378995/confused-with-java-overriding-the-access-level – Sotirios Delimanolis Oct 16 '15 at 16:01

score 1 · Answer 2 · answered Oct 16 '15 at 13:29

In Java, private methods are NOT visible to subclasses. Protected methods ARE.

In the class Two you did not override the start() method, therefore t.start() uses One.start() which calls One.run() which prints "one".

If you change the run() method to public or protected the Two class will override it and calling t.start() will end up calling the version declared in Two.

In "flowchart" form:
When run is private

Two.start() -> not defined! -> go to superclass -> One.start() -> run() -> One.run()

When run is public or protected

Two.start() -> not defined! -> go to superclass -> One.start() -> run() -> run() is overridden by subclass -> go to subclass -> Two.run()

score 0 · Answer 3 · answered Oct 16 '15 at 13:27

0

Because run is private in One, it is not dynamically linked, nor inherited. The run in Two does NOT override the run in One

answered Oct 16 '15 at 13:27

ControlAltDel

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Inheritance and public/private methods

3 Answers3