Java generic method boundaries

Question

What is the difference in following 2 lines?

public static <T extends Comparable<? super T>> int methodX(List<T> data)
public static <T> int methodX(List<? extends Comparable<? super T>> data)

Answer 1

Your first option is a "stricter" parametrisation. Meaning, you're defining the class T with a bunch of restrictions, and then use it later on with List. In your second method, the parameter class T is generic with no conditions, and the Lists class parameter is defined in terms of the parameter T.

The second way is syntactically different as well, with a ? instead of the first option's T, because in the parameter definition you aren't defining the type parameter T but rather using it, so the second method cannot be as specific.

The practical difference that comes out of this is one of inheritance. Your first method needs to be a type that is comparable to a super class of itself, whereas the second type need only be comparable to an unconditional/unrelated T:

public class Person implements Comparable<Number> {
    @Override
    public int compareTo(Number o) {
        return 0;
    }
    public static <T extends Comparable<? super T>> int methodX(List<T> data) {
            return 0;
    }
    public static <T> int methodY(List<? extends Comparable<? super T>> data) {
            return 0;
    }
    public static void main(String[] args) {
        methodX(new ArrayList<Person>()); // stricter ==> compilation error
        methodY<Object>(new ArrayList<Person>());
    }
}

If you change the Comparable of Person to be able to compare Object or Person (the inheritance tree of the base class) then methodX will also work.