Invoke non-abstract method of super interface in concrete class

Question

The Calculator interface has calculate abstract method and ramdom() non-abstract method. I want to use super ramdom() and also Override ramdom() at concrete class CalculatorImpl. My question is why I have to call like that Calculator.super.ramdom() ? Why super.ramdon() don't work?

public interface Calculator {

    double calculate(int a);

    default double ramdom() {
        System.out.println("Calculator ramdom");
        return Math.random();
    }

}

class CalculatorImpl implements Calculator {

    @Override
    public double calculate(int a) {
        // calling super.ramdom() will get [The method ramdom() is undefined for
        // the type Object error]
        return Calculator.super.ramdom() * a;
    }

    @Override
    public double ramdom() {
        System.out.println("CalculatorImpl ramdom");
        return 0;
    }

}

Possible duplicate of [How to spell random](http://www.spellweb.com/how-do-you-spell/random) — shmosel, Apr 04 '16 at 07:23

score 2 · Accepted Answer · answered Apr 04 '16 at 07:09

To answer your exact question: when you use super inside a class ... you are "pointing" to the class it is extending; in this case Object.

But Object does not have a method random that you could be calling.

Therefore you have to make it explicit "where" random is actually coming from.

score 0 · Answer 2 · edited May 23 '17 at 12:32

0

Just remove the override of ramdom in CalculatorImpl. It will implicitly used the default implementation from Calculator.

class CalculatorImpl implements Calculator {

@Override
public double calculate(int a) {
    // calling super.ramdom() will get [The method ramdom() is undefined for
    // the type Object error]
    return Calculator.super.ramdom() * a;
}
/* Remove the Override and the default implementation will be used
@Override
public double ramdom() {
    System.out.println("CalculatorImpl ramdom");
    return 0;
}
*/
}

Keyword super is used to call the inherited class and not the implemented interface. But here you have no inheritance except the implicit Object which does not have a random method

Edit: I might have missunderstood your question. If you want to override and call the default implementation, check the link provided by Sasha Salauyou

edited May 23 '17 at 12:32

Community

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answered Apr 04 '16 at 07:07

ortis

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in fact, removing `@Override` will not affect anything. It is just an annotation telling compiler you're gonna override some method (so it could check the signature and tell if you're mistaken) – Alex Salauyou Apr 04 '16 at 07:15
@Sasha Salauyou I am certainly not talking about the annotation but the whole method, as the code sample shows ;) – ortis Apr 04 '16 at 07:16
1

I see, but it also has no sense. OP says they need overriden method as well. – Alex Salauyou Apr 04 '16 at 07:17
@Sasha Salauyou you're right. I will be editing this post – ortis Apr 04 '16 at 07:20
@SashaSalauyou my question is why I have to call explicitly ? I know how to call a default method explicitly – Mizuki Apr 04 '16 at 15:41
@ortis my question is why I have to call explicitly ? I know how to call a default method explicitly – Mizuki Apr 04 '16 at 15:42
@Alice the answer is: this is rule of the language – Alex Salauyou Apr 04 '16 at 15:52

Invoke non-abstract method of super interface in concrete class

2 Answers2