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

For objects of type X to be comparable with each other, class X has to implement exactly Comparable.

This is not what your code is doing, you've got a class Item and you are implementing Comparable instead of Comparable>. This means that Item can be compared with T, but not with Item, which is required.

Change your Item class to:

public abstract class Item implements Comparable>
{
    protected T item;

    @Override
    public int compareTo(Item o)
    {
        return 0; // this doesn't matter for the time being
    }
}