understanding Comparable interface in Java

Question

I have a problem with understanding Comparable interface in Java. Here is a code:

public class test{
public class Example1 implements Comparable<Example1> {
    private int id;

    public Example1(int id) {
        this.id = id;
    }

    public int getId() {
        return this.id;
    }

    public void setId(int id) {
        this.id = id;
    }


    @Override
    public int compareTo(Example1 o) {
        if (id == o.getId())
            return 0;
        else if (id > o.getId())
            return 1;
        else
            return -1;
    }
}
public class Example2 extends Example1 implements Comparable<Example2> {
    public Example2(int id) {
        super(id);
    }
}

}

When I want to implement Comparable a second time I get an error java.lang.Comparable' cannot be inherited with different type arguments.

Why I'm not allowed to to implement a child class with Comparable interface? How can I resolve this problem? I will be very thankful for answer.

You can't declare a different comparable because all `Example1`s (and subclasses) already implement `Comparable`, and thus a subclass can't implement a different `Comparable` as well. It's the same reason you can't make a single class implement `Comparable` twice. — Andy Turner, Apr 17 '21 at 11:24
I am not sure, but try put T as Example 1 parameter and give it to comparable. When you extend Example1 then, give Example2 — DynoZ, Apr 17 '21 at 11:39
public class A implements Comparable{} then public class B extends A — DynoZ, Apr 17 '21 at 11:40
I agree that it’s a weird restriction of Java generics, but it’s a restriction. The explanation for it can be found in the way generics are implemented. — Ole V.V., Apr 17 '21 at 11:47
A tip: Paste your error message into your search engine. That will very often lead to both an explanation and a solution. — Ole V.V., Apr 17 '21 at 11:47

understanding Comparable interface in Java

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