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

Question

Consider the code:

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

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

public class MyItem<T> extends Item<String>
{
    T object;
}

public class Foo<T>
{
    protected ArrayList<T> list;
}

public class Bar<V> extends Foo<MyItem<V>>
{
    public void sort()
    {
        Collections.sort(list);
    }
}

The sort call gives the error:

Bound mismatch: The generic method sort(List< T >) of type Collections is not applicable for the arguments (ArrayList< MyItem< T > >). The inferred type MyItem< T > is not a valid substitute for the bounded parameter < T extends Comparable< ? super T > >

Why is this wrong?

If MyItem<V> implements Comparable then why is it not a substitute?

Sorry if this has been asked, but I feel the question is somewhat specific.

score 12 · Accepted Answer · edited Sep 28 '13 at 10:50

Actually more detailed explanation of this error gives your javac itself:

java: no suitable method found for sort(java.util.ArrayList<MyItem<V>>)

method java.util.Collections.<T>sort(java.util.List<T>,java.util.Comparator<? super T>) is not applicable (cannot instantiate from arguments because actual and formal argument lists differ in length)

method java.util.Collections.<T>sort(java.util.List<T>) is not applicable (inferred type does not conform to declared bound(s) inferred: MyItem<V> bound(s): java.lang.Comparable<? super MyItem<V>>)

So, the main question is:
why is method Collections.<T>sort(java.util.List<T>)) not applicable?

The answer is:
because in Collections.<T>sort(java.util.List<T>) method declaration there are bounds on parameter T: <T extends Comparable<? super T>>.

In another words, T must implement Comparable interface on it self. For example String class implements such interface: ...implements ... Comparable<String>.

In your case Item class doesn't implement such interface:

Item<T> implements Comparable<T> is not same thing as Item<T> implements Comparable<Item<T>>.

So, for solving this problem, your should change your Item class to this one:

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

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

That was spot on. Cheers! For clarification, I included the extra classes because the code I provided is a much more simplified version of my actual code, and I wasn't sure where the problem was. I completely missed that I could just change the actual comparison itself. Thanks again. — terrorcell, Jul 19 '13 at 08:00
Better : `public abstract class Item implements Comparable extends Item>` - no ? — Mr_and_Mrs_D, Sep 28 '13 at 10:50

score 2 · Answer 2 · answered Jul 19 '13 at 07:31

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

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

Change your Item<T> class to:

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

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

"For objects of type X to be comparable with each other, class X has to implement exactly Comparable." Not necessarily exactly. X could implement `Comparable` — newacct, Sep 30 '13 at 23:59

score 1 · Answer 3 · answered Jul 19 '13 at 07:30

Just change the class like follow:

     public class MyItem<T> extends Item<String> implement Comparable<MyItem<T>>
     {
         T object;
     }

Or

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

           public int compareTo(MyItem<T> o)
           {
              return 0; // this doesn't matter for the time being
       }

}

The error tips has shown us.Hope it helpful.

score 0 · Answer 4 · answered Jul 19 '13 at 07:25

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<String> {}

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<String>, 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<T> {
    protected T item;
}

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

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

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

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