Inferred type is not a valid substitute for a Comparable generic type

Question

Consider the code:

public abstract class Item implements Comparable
{
    protected T item;

    public int compareTo(T o)
    {
        re         


        

Answer 1

You do not need to have the class MyItem generified just to see the effect. The following class is enough to see what happens:

public class MyItem extends Item {}

Now you have the following call:

Collections.sort(list);

As morgano stated correctly, the sort method will take a collection that is parameterized with a type T that must be comparable to T. Your MyItem class is extending Item, which results in MyItem being comparable to Strings.

With a little switch in which class implements the Comparable interface, you will get the expected result:

public abstract class Item {
    protected T item;
}

public class MyItem extends Item implements Comparable {
    @Override
    public int compareTo(MyItem o) {
        return item.compareTo(o.item); // just an example
    }
}

And now the call to Collections.sort(list) will work.