Question

//src/com/test/animal/Animal.java
package com.test.animal;

public class Animal
{
    Animal()
    {
        init();
    }

    public void init()
    {
        System.out.println("parent init()");
    }
}

//src/com/test/animal/Dog.java 
package com.test.animal;

public class Dog extends Animal
{
    String name = null;

    Dog()
    {
        super();
    }

    public void init()
    {
        System.out.println("child init()");
        super.init();
        name = new String("dog");
        System.out.println("name: "+name);
    }

    public static void main(String[] args)
    {
        Dog d = new Dog();
        System.out.println("name: "+d.name);
    }
}

The output is:

child init()
parent init()
name: dog
name: null

It seems the init() in child is called, but NAME value not saved! Why?
It would be OK if I move NAME to parent. However, it’s more resonable to retain in child, since it’s Dog-specific.

Also, I can explicitly call init() in child’s constructor to solve this issue. It’s not that good.

Answer 1

The order in which this executes is as follows:

1. The Dog constructor is called.
2. It calls the Animal constructor.
3. The Animal constructor calls the init method. Because it is overridden in Dog, the Dog version is called.
4. In Dog.init you set name to "dog".
5. The Dog.init method returns.
6. Now, the member variables of Dog are initialized. This sets name to null.

Result: name will be null.

An excellent example of why you should not call methods that can be overridden from a constructor – because it leads to surprises like this.

Side note: Never do this:

// Unnecessarily creating a new String object
name = new String("dog");

Just do this instead:

name = "Dog";

It is never necessary to explicitly create a new String object from a string literal.