Java lang unsupportedoperationexception ошибка

I have this code:

public static String SelectRandomFromTemplate(String template,int count) {
   String[] split = template.split("|");
   List<String> list=Arrays.asList(split);
   Random r = new Random();
   while( list.size() > count ) {
      list.remove(r.nextInt(list.size()));
   }
   return StringUtils.join(list, ", ");
}

I get this:

06-03 15:05:29.614: ERROR/AndroidRuntime(7737): java.lang.UnsupportedOperationException
06-03 15:05:29.614: ERROR/AndroidRuntime(7737):     at java.util.AbstractList.remove(AbstractList.java:645)

How would be this the correct way? Java.15

I think that replacing:

List<String> list = Arrays.asList(split);

with

List<String> list = new ArrayList<String>(Arrays.asList(split));

resolves the problem.

Just read the JavaDoc for the asList method:

Returns a {@code List} of the objects
in the specified array. The size of
the {@code List} cannot be modified,
i.e. adding and removing are
unsupported, but the elements can be
set. Setting an element modifies the
underlying array.

This is from Java 6 but it looks like it is the same for the android java.

EDIT

The type of the resulting list is Arrays.ArrayList, which is a private class inside Arrays.class. Practically speaking, it is nothing but a List-view on the array that you’ve passed with Arrays.asList. With a consequence: if you change the array, the list is changed too. And because an array is not resizeable, remove and add operation must be unsupported.

answered Jun 3, 2010 at 12:14

answered Jun 3, 2010 at 13:30

I’ve got another solution for that problem:

List<String> list = Arrays.asList(split);
List<String> newList = new ArrayList<>(list);

work on newList

Replace

List<String> list=Arrays.asList(split);

to

List<String> list = New ArrayList<>();
list.addAll(Arrays.asList(split));

or

List<String> list = new ArrayList<>(Arrays.asList(split));

or

List<String> list = new ArrayList<String>(Arrays.asList(split));

or (Better for Remove elements)

List<String> list = new LinkedList<>(Arrays.asList(split));

answered Nov 16, 2018 at 13:56

Yes, on Arrays.asList, returning a fixed-size list.

Other than using a linked list, simply use addAll method list.

Example:

String idList = "123,222,333,444";

List<String> parentRecepeIdList = new ArrayList<String>();

parentRecepeIdList.addAll(Arrays.asList(idList.split(","))); 

parentRecepeIdList.add("555");

You can’t remove, nor can you add to a fixed-size-list of Arrays.

But you can create your sublist from that list.

list = list.subList(0, list.size() - (list.size() - count));

public static String SelectRandomFromTemplate(String template, int count) {
   String[] split = template.split("\|");
   List<String> list = Arrays.asList(split);
   Random r = new Random();
   while( list.size() > count ) {
      list = list.subList(0, list.size() - (list.size() - count));
   }
   return StringUtils.join(list, ", ");
}

*Other way is

ArrayList<String> al = new ArrayList<String>(Arrays.asList(template));

this will create ArrayList which is not fixed size like Arrays.asList

Arrays.asList() uses fixed size array internally.
You can’t dynamically add or remove from thisArrays.asList()

Use this

Arraylist<String> narraylist=new ArrayList(Arrays.asList());

In narraylist you can easily add or remove items.

Arraylist narraylist=Arrays.asList(); // Returns immutable arraylist
To make it mutable solution would be:
Arraylist narraylist=new ArrayList(Arrays.asList());

Following is snippet of code from Arrays

public static <T> List<T> asList(T... a) {
        return new ArrayList<>(a);
    }

    /**
     * @serial include
     */
    private static class ArrayList<E> extends AbstractList<E>
        implements RandomAccess, java.io.Serializable
    {
        private static final long serialVersionUID = -2764017481108945198L;
        private final E[] a;

so what happens is that when asList method is called then it returns list of its own private static class version which does not override add funcion from AbstractList to store element in array. So by default add method in abstract list throws exception.

So it is not regular array list.

answered Apr 15, 2016 at 4:10

Creating a new list and populating valid values in new list worked for me.

Code throwing error —

List<String> list = new ArrayList<>();
   for (String s: list) {
     if(s is null or blank) {
        list.remove(s);
     }
   }
desiredObject.setValue(list);

After fix —

 List<String> list = new ArrayList<>();
 List<String> newList= new ArrayList<>();
 for (String s: list) {
   if(s is null or blank) {
      continue;
   }
   newList.add(s);
 }
 desiredObject.setValue(newList);

answered Sep 27, 2019 at 10:29