Java- why does it print null?

Question

public class Base {
public Base() {
    x = 0;
    bar();
}

public Base(int x) {
    this.x = x;
    foo();
}

public void foo() {
    System.out.println("Base.foo : " + x);
}

private void bar() {
    System.out.println("Base.bar:" + x.toString());
}

protected Integer x;
}

    public class Derived extends Base {
       public Derived() {
       bar();
       }
       public Derived(int x, int y) {
         super(x);
         this.y = y;
       }
       public void foo() {
         System.out.println("Derived.foo : " + x + ", " + y);
       }
       public void bar() {
         System.out.println("Derived.bar:" + x.toString() + ", " + y.toString());
       }
       private Integer y;


       public static void main(String[] args) {
        Base b = new Derived(10, 20);
      }
}

Why does it print "Derived.foo:" 10, nulll and not 20 instead of the null? y is a private variable of Derived, and it was initialized with 20. it's in its scope.. so why is it null?

Answer 1

Because the super constructor is first called (super(x)). This super constructor calls the method foo. Then the Derived constructor initializes y to 20 (this.y = y). So when foo is called, y is not initialized yet.

It's a bad practice to call overridable methods in constructors, because, as you just noticed, it can call overridden methods on partially constructed objects.

Answer 2

The println comes from method foo which is called from Base's constructor, which is called from Derived's constructor (via super) before you initialize y. So the null value is expected.