Java - extends why the super variable a is 0

Question

Please look at this code:

class Sup {
    int a = 8;

    public void printA() {
        System.out.println(a);
    }

    Sup() {
        printA();
    }
}

public class Sub extends Sup {
    int a = 9;

    @Override
    public void printA() {
        System.out.println(a);
    }

    Sub() {
        printA();
    }

    public static void main(String[] args) {
        Sub sub = new Sub();
    }
}

result： console print： 0 9
I know that subclass will first calls the superclass constructor
but ,why is the 0 9 , not 8 9?

Are you sure the constructor is called in the order you think it is? — OneCricketeer, Aug 16 '17 at 13:11
@cricket_007: Of course it's called. You **cannot** avoid calling it in Java. — T.J. Crowder, Aug 16 '17 at 13:12
@T.J.Crowder You can avoid it by using serialization or cloning. — Flown, Aug 16 '17 at 13:19
@Flown: Meh, the super constructor's been called at *some point* in both those cases. :-D — T.J. Crowder, Aug 16 '17 at 13:25
@T.J.Crowder You're right in terms of the already existing objects. — Flown, Aug 16 '17 at 14:34

score 14 · Accepted Answer · answered Aug 16 '17 at 13:11

14

When the Sup constructor calls printA() it executes the printA method of class Sub (which overrides the method of the same name of class Sup), so it returns the value of the a variable of class Sub, which is still 0, since the instance variables of Sub are not yet initialized (they are only initialized after the Sup constructor is done).

answered Aug 16 '17 at 13:11

Eran

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This is also why some code standards require all methods called in constructors to be `final` or `private`. To prevent overrides causing these situations. – Kiskae Aug 16 '17 at 13:26

Java - extends why the super variable a is 0

1 Answers1