resolve a java.util.ArrayList$SubList notSerializable Exception

Question

I am using SubList function on an object of type List. The problem is that I am using RMI and because the java.util.ArrayList$SubList is implemented by a non-serializable class I got the Exception described above when I try to pass the resulting object to a remote function taking as an argument a List as well. I've seen that I should copy the resulting List to a new LinkedList or ArrayList and pass that.

Does anyone know a function that helps as to easily do that for this for example ?

List<String> list = originalList.subList(0, 10);

You can copy a list very easily. The ArrayList constructor will accept an existing collection (including your sublist) as an argument, and perform a copy. — Afforess, Oct 25 '14 at 23:15
@Afforess stole my answer. Just copy it to a list you know is serializable. — markspace, Oct 25 '14 at 23:46
@Afforess Why don't you post your comment as an answer? Then the OP can accept your answer. — eee, Oct 26 '14 at 10:31

aravindaM · Accepted Answer · 2014-10-26T17:57:25.903

53

It's because, List returned by subList() method is an instance of 'RandomAccessSubList' which is not serializable. Therefore you need to create a new ArrayList object from the list returned by the subList().

ArrayList<String> list = new ArrayList<String>(originalList.subList(0, 10));

edited Oct 26 '14 at 17:57

answered Oct 26 '14 at 16:35

aravindaM

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score 7 · Answer 2 · answered Oct 26 '14 at 16:17

7

The solution was simply this code:

ArrayList<String> list = new ArrayList<String>();
list.addAll(originalList.subList(0, 10));

answered Oct 26 '14 at 16:17

Othmane

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The other proposed solution is better because it will pre-allocate the full size once, instead of possibly having to allocate the default size & then re-allocate space for the sublist while adding the elements. – Daniel Gray Nov 18 '16 at 13:31
If you care about performance you should not create twice the same sub-list – Jack Nov 16 '20 at 10:45

resolve a java.util.ArrayList$SubList notSerializable Exception

2 Answers2