Does a class which implements an interface's method (without explicitly implementing that interface) extend that specific interface?

Question

I am implementing a class to store objects that can be assigned a double value. For this reason, I have created a HasDoubleValue interface, that contains a single method:

public interface HasDoubleValue{
    public double doubleValue();
}

My main class is defined as such:

Data <O extends HasDoubleValue> {...}

Now, when I try to initialize this class to store Integers, I get a "type argument Integer is not within bounds of type-variable O" error, although Integer implements the doubleValue() method by default.

I suppose that this happens because Integer does not explicitly implement my HasDoubleValue interface, although it has the method I am looking for. Is this right? What would a natural workaround be?

Answer 1

Yes, it is right. Java doesn't use duck-typing as JavaScript or TypeScript.

A solution is to create an adapter class that wraps a Integer, delegates to it, and actually implement the interface.

Or, since inthis case your interface is a functional interface, to use a lambda or a method reference to create an instance of HasDoubleValue from an Integer.

public interface HasDoubleValue{
    double doubleValue();
}

final class IntegerHasDoubleValueAdapter implements HasDoubleValue {
    private final Integer i;

    public IntegerHasDoubleValueAdapter(Integer i) {
        this.i = i;
    }

    @Override
    public double doubleValue() {
        return i.doubleValue();
    }
}

class Data<O extends HasDoubleValue> {
    void put(O o) {}

    public static void main(String[] args) {
        Integer i = 42;

        Data<IntegerHasDoubleValueAdapter> d1 = new Data<>();
        d1.put(new IntegerHasDoubleValueAdapter(i));

        Data<HasDoubleValue> d2 = new Data<>();
        d2.put(() -> i.doubleValue());

        Data<HasDoubleValue> d3 = new Data<>();
        d3.put(i::doubleValue);
    }
}