Java接口扩展了Comparable

Question

I want to have an interface A parameterised by T A<T> , and also want every class that implements it to also implement Comparable (with T and its subtypes). It would seem natural to write interface A<T> extends Comparable<? extends T> interface A<T> extends Comparable<? extends T> , but that doesn't work. How should I do it then?

Answer 1

When Comparable<? extends T> Comparable<? extends T> appears it means you have an instance of Comparable that can be compared to one (unknown) subtype of T , not that it can be compared to any subtype of T.

But you don't need that, because a Comparable<T> can compare itself to any subtype of T anyway, eg a Comparable<Number> can compare itself to a Comparable<Double> .

So try:

interface A<T> extends Comparable<T> {
    // ...
}

or

interface A<T extends Comparable<T>> extends Comparable<A<T>> {
    // ...
}

depending on whether you need to be able to compare instances of T in order to implement your compareTo method.

Answer 2

If you use comparable you do not need to specify the possibility for subtypes in the compare function, it is by nature possible to pass in any subtype of an object X into a method that declared a parameter of class X. See the code below for more information.

public interface Test<T> extends Comparable<T> {

}

class TestImpl implements Test<Number> {
    @Override
    public int compareTo(final Number other) {
        return other.intValue() - 128;
    }
}

class TestMain {
    public static void main(final String[] args) {
        TestImpl testImpl = new TestImpl();
        testImpl.compareTo(Integer.MIN_VALUE);
    }
}