Java Puzzle - initialization order

Overview

When I applied my first full-time job, a guy during recruitment showed me the following code and asked me what would happen? It supposed to show if I understand initialization order in Java.

Example

class ClassMain {

    private final String mainVar = "Main";

    ClassMain() {
        showMeTheMain();
    }

    protected void showMeTheMain() {
        System.out.println(mainVar);
    }
}

class ClassA extends ClassMain {

    private String a = "qwerty";

    ClassA() {
        showMeTheMain();
    }

    @Override
    public void showMeTheMain() {
        System.out.println("not main anymore " + a);
    }
}

Solution

Original question had System.out.println("not main anymore " + a.toUpperCase());, but example in this post is more interesting.

The result is:

not main anymore null
not main anymore qwerty

Why? In the constructor of base class we try to execute overridden version of showMeTheMain, but a is not initialized yet.

This is how program is executed:

• Constructor of ClassMain
• Method ClassA.showMeTheMain
• Constructor of ClassA (where a is initialized)

private vs private final

If we change private String a = "qwerty"; to private final String a = "qwerty"; we get:

not main anymore qwerty
not main anymore qwerty

private vs private final and initialize in constructor.

If we change private String a = "qwerty"; to private final String a; and initialize in the constructor we should get again:

not main anymore null
not main anymore qwerty

Note: In this example we changed access qualification from protected to public. We can increase access qualification, but we cannot make it more restrict.