Why can't I use ArrayList<String> as a parameter for a generic ArrayList<E> method?

Question

I am learning about generics so I made a simple java program so I can learn how they exactly work. But now I am stuck with this program because I can't see why it won't work.

public static void main(String[] args)
    {
        ArrayList<String> strings = new ArrayList<>();
        strings.add("A");
        strings.add("B");

        printList(strings);
    }

    public static <E extends List<E>> void printList(ArrayList<E> l)
    {
        for (E obj : l)
        {
            System.out.println(obj.toString());
        }
    }

Because you've specified E to extend List<E>. Which is not true for String. — lexicore
– lexicore, Commented Jun 18, 2018 at 18:33
Your formal argument is effectively ArrayList<? extends List<E>>. Either replace the parameter type with E, or drop the bound on E. — Mike Strobel
– Mike Strobel, Commented Jun 18, 2018 at 18:33
What you've specified with your method generics is that the argument passed to the method will be an ArrayList<List<E>>. A list of lists, basically. Then you pass it a list of String. Just take out the <E extends List<E>> part and you should be good to go. — mypetlion
– mypetlion, Commented Jun 18, 2018 at 18:36

Louis Wasserman · Accepted Answer · 2018-06-18 18:33:11Z

6

<E extends List<E>> is almost certainly not what you want. What that says is that your element type is a list of itself.

Instead, write just <E>.

answered Jun 18, 2018 at 18:33

Louis Wasserman

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Comments

score 5 · Accepted Answer · 2018-06-18 18:53:01Z

5

Your generic parameter E should not extend list. if E extends List<E> then you should be expecting List< List<> >.

Your method should look something like:

public static <E> void printList(List<E> list) {

    for (E obj : list) {

        System.out.println(obj);
    }
}

Notes:

Use meaningful argument names like list not l so that code maintainers understand your code
System.out.println(object.toString()) can be simplified to System.out.println(object) because toString() will be called automatically
If you want you code to be more robust, make the parameter a List<E> so you can accept all List<E> objects such as LinkedList<E> and all other implementations.

edited Jun 18, 2018 at 18:53

answered Jun 18, 2018 at 18:34

user5063151

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