Inheritance issue that i cannot understand

Question

I have this 2 classes:

public class A {
    protected int _x;

    public A() {
        _x = 1;
    }

    public A(int x) {
        _x = x;
    }

    public void f(int x) {
        _x += x;
    }

    public String toString() {
        return "" + _x;
    }
}

public class B extends A {
    public B() {
        super(3);
    }

    public B(int x) {
        super.f(x);
        f(x);
    }

    public void f(int x) {
        _x -= x;
        super.f(x);
    }

    public static void main(String[] args) {
        A[] arr = new A[3];
        arr[0] = new B();
        arr[1] = new A();
        arr[2] = new B(5);
        for (int i = 0; i < arr.length; i++) {
            arr[i].f(2);
            System.out.print(arr[i] + " ");
        }
    }
}

The output is 3 3 6 and I am wonder why the third iteration is 6

Would you mind making the whole example somewhat reasonble? I do not mind reading code that itself is quite mind-numbling, however reading half-obfuscated variable/class names gets annoying quickly. Also add @Override tags. — skiwi
– skiwi, Commented Feb 8, 2014 at 15:04
Just go through the code in your mind... When you do not call a super constructor, the empty super constructor is called. — Matthias
– Matthias, Commented Feb 8, 2014 at 15:05
Write each step on paper. Remember that the first thing a constructor does if it doesn't explicitely invoke super() is to invoke the super constructor without argument. — JB Nizet
– JB Nizet, Commented Feb 8, 2014 at 15:05
@JB Nizet, except you call some other super constructor with arguments. — Matthias
– Matthias, Commented Feb 8, 2014 at 15:06

Rohit Jain · Accepted Answer · 2014-02-08 15:05:46Z

6

The constructor:

public B(int x)
{
    super.f(x);
    f(x);
}

is translated by the compiler to this:

public B(int x)
{
    super();
    super.f(x);
    f(x);
}

I guess now you would understand, why it's 6.

answered Feb 8, 2014 at 15:05

Rohit Jain

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4 Comments

user3271698 Over a year ago

after the first super _x = 1, and after that i have super.f(x) so it goes to f function inside A class but because i have the function inside B class it implement f from B class and inside this f i have another time super.f(x) so it looks like recursion

Rohit Jain Over a year ago

@user3271698 super.f(x) won't call the overridden method.

user3271698 Over a year ago

Why not ? its B object

JB Nizet Over a year ago

Because super.f(x) precisely means: call the superclass implementation of the f() method.

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