Question

I wrote a function to process some ArrayList, and want to pass out the lists to outer caller. But I found the list was not changed after calling the function.

Here are some test code (of coz, it’s a demo but illustrates my problem):

@Test
public void testList() {
    List<Integer> l1 = new ArrayList<Integer>(5);
    List<Integer> l2 = new ArrayList<Integer>(5);
    l1.add(11); l1.add(12); l1.add(13); l1.add(14); l1.add(15);
    l2.add(21); l2.add(22); l2.add(23); l2.add(24); l2.add(25);
    System.out.println("outer l1: " + l1);
    System.out.println("outer l2: " + l2);

    subList(l1, l2);  //call the function

    System.out.println("after l1: " + l1);
    System.out.println("after l2: " + l2);
}

protected static void subList(List<Integer> l1, List<Integer> l2) {
    l1.remove(0);  l1.remove(0);
    l2 = l2.subList(0, 4);
    System.out.println("inner l1: " + l1);
    System.out.println("inner l2: " + l2);
}

And the result is:

outer l1: [11, 12, 13, 14, 15]
outer l2: [21, 22, 23, 24, 25]
inner l1: [13, 14, 15]             (in function, changed)
inner l2: [21, 22, 23, 24]         (in function, changed too)
after l1: [13, 14, 15]             (changed)
after l2: [21, 22, 23, 24, 25]     (unchanged?!)

It seems that List2 has not changed by function subList(List l1, List l2), while List1 changed.

Is the List parameter passed by CopyValue?? So, in the function, when l2’s reference changed, JVM copy a new list, the original outer list is maintain old value?

How to write a “side-effect” procedure to modify ArrayList ?

Answer 1

References are always passed by value in Java. So, assigning a new value to l2 in the called method won’t change the reference in the calling method.

You need the following code for it to work:

List<Integer> subList = new ArrayList<Integer>(l2.subList(0, 4));
l2.clear();
l2.addAll(subList);