Difference between Arrays.asList(array) and new ArrayList<Integer>(Arrays.asList(array))

Question

What is the difference between

• List<Integer> list1 = new ArrayList<Integer>(Arrays.asList(ia)); // Copy

• List<Integer> list2 = Arrays.asList(ia);

, where ia is an array of integers?

I came to know that some operations are not allowed in list2. Why is it so? How is it stored in memory (references / copy)?

When I shuffle the lists, list1 doesn't affect the original array, but list2 does. But still list2 is somewhat confusing.

How does ArrayList being upcasted to list differ from creating a new ArrayList?

list1 differs from (1)
ArrayList<Integer> list1 = new ArrayList<Integer>(Arrays.asList(ia));

Answer 1

First of all, the Arrays class is a utility class which contains a number of utility methods to operate on Arrays (thanks to the Arrays class. Otherwise, we would have needed to create our own methods to act on Array objects)

asList() method:

1. asList method is one of the utility methods of Array class, it is a static method that's why we can call this method by its class name (like Arrays.asList(T...a) )
2. Now here is the twist. Please note that this method doesn't create new ArrayList object. It just returns a List reference to an existing Array object (so now after using asList method, two references to existing Array object gets created)
3. and this is the reason. All methods that operate on List object, may not work on this Array object using the List reference. Like for example, Arrays size is fixed in length, hence you obviously can not add or remove elements from Array object using this List reference (like list.add(10) or list.remove(10);. Else it will throw UnsupportedOperationException).
4. any change you are doing using a list reference will be reflected in the exiting Arrays object (as you are operating on an existing Array object by using a list reference)

In the first case, you are creating a new Arraylist object (in the second case, only a reference to existing Array object is created, but not a new ArrayList object), so now there are two different objects. One is the Array object and another is the ArrayList object and there isn't any connection between them (so changes in one object will not be reflected/affected in another object (that is, in case 2, Array and Arraylist are two different objects)

Case 1:

Integer [] ia = {1,2,3,4};
System.out.println("Array : "+Arrays.toString(ia));
List<Integer> list1 = new ArrayList<Integer>(Arrays.asList(ia));  // new ArrayList object is created , no connection between existing Array Object
list1.add(5);
list1.add(6);
list1.remove(0);
list1.remove(0);
System.out.println("list1: " + list1);
System.out.println("Array: " + Arrays.toString(ia));

Case 2:

Integer [] ia = {1,2,3,4};
System.out.println("Array: " + Arrays.toString(ia));
List<Integer> list2 = Arrays.asList(ia); // Creates only a (new) List reference to the existing Array object (and NOT a new ArrayList Object)
//  list2.add(5); // It will throw java.lang.UnsupportedOperationException - invalid operation (as Array size is fixed)
list2.set(0,10);  // Making changes in the existing Array object using the List reference - valid 
list2.set(1,11); 
ia[2]=12;     // Making changes in the existing Array object using the Array reference - valid
System.out.println("list2: " + list2);
System.out.println("Array: " + Arrays.toString(ia));

Answer 2

Many people have answered the mechanical details already, but it's worth noting: This is a poor design choice, by Java.

Java's asList method is documented as "Returns a fixed-size list...". If you take its result and call (say) the .add method, it throws an UnsupportedOperationException. This is unintuitive behavior! If a method says it returns a List, the standard expectation is that it returns an object which supports the methods of interface List. A developer shouldn't have to memorize which of the umpteen util.List methods create Lists that don't actually support all the List methods.

If they had named the method asImmutableList, it would make sense. Or if they just had the method return an actual List (and copy the backing array), it would make sense. They decided to favor both runtime-performance and short names, at the expense of violating both the principle of least astonishment and the good object-oriented practice of avoiding UnsupportedOperationExceptions.

(Also, the designers might have made a interface ImmutableList, to avoid a plethora of UnsupportedOperationExceptions.)

Answer 3

String names[] = new String[]{"Avinash","Amol","John","Peter"};
java.util.List<String> namesList = Arrays.asList(names);

or

String names[] = new String[]{"Avinash","Amol","John","Peter"};
java.util.List<String> temp = Arrays.asList(names);

The above statement adds the wrapper on the input array. So the methods like add and remove will not be applicable on the list reference object 'namesList'.

If you try to add an element in the existing array/list then you will get "Exception in thread "main" java.lang.UnsupportedOperationException".

The above operation is readonly or viewonly.
We can not perform add or remove operation in list object.

But

String names[] = new String[]{"Avinash","Amol","John","Peter"};
java.util.ArrayList<String> list1 = new ArrayList<>(Arrays.asList(names));

or

String names[] = new String[]{"Avinash","Amol","John","Peter"};
java.util.List<String> listObject = Arrays.asList(names);
java.util.ArrayList<String> list1 = new ArrayList<>(listObject);

In the above statement you have created a concrete instance of an ArrayList class and passed a list as a parameter.

In this case, methods add and remove will work properly as both methods are from ArrayList class, so here we won't get any UnSupportedOperationException. Changes made in the Arraylist object (method add or remove an element in/from an arraylist) will get not reflect in to the original java.util.List object.

String names[] = new String[] {
    "Avinash",
    "Amol",
    "John",
    "Peter"
};

java.util.List < String > listObject = Arrays.asList(names);
java.util.ArrayList < String > list1 = new ArrayList < > (listObject);
for (String string: list1) {
    System.out.print("   " + string);
}
list1.add("Alex"); // Added without any exception
list1.remove("Avinash"); // Added without any exception will not make any changes in original list in this case temp object.


for (String string: list1) {
    System.out.print("   " + string);
}
String existingNames[] = new String[] {
    "Avinash",
    "Amol",
    "John",
    "Peter"
};
java.util.List < String > namesList = Arrays.asList(names);
namesList.add("Bob"); // UnsupportedOperationException occur
namesList.remove("Avinash"); // UnsupportedOperationException

Answer 4

Arrays.asList()

This method returns its own implementation of List. It takes an array as an argument and builds methods and attributes on top of it, since it is not copying any data from an array but using the original array this causes alteration in original array when you modify list returned by the Arrays.asList() method.

On the other hand, ArrayList(Arrays.asList()); is a constructor of ArrayList class which takes a list as argument and returns an ArrayList that is independent of list, i.e., Arrays.asList() in this case passed as an argument.

That is why you see these results.

Answer 5

An explanation with documentation references would be better for someone looking for answer.

1. java.util.Arrays

• This is a utility class with bunch of static methods to operate on given array
• asList is one such static method that takes input array and returns an object of java.util.Arrays.ArrayList which is a static nested class that extends AbstractList<E> which in turn implements List interface.
• So Arrays.asList(inarray) returns a List wrapper around the input array, but this wrapper is java.util.Arrays.ArrayList and not java.util.ArrayList and it refers to the same array, so adding more elements to the List wrapped array would affect the original one too and also we cannot change the length.

2. java.util.ArrayList

• ArrayList has a bunch of overloaded constructors

   public ArrayList() - // Returns arraylist with default capacity 10
  
   public ArrayList(Collection<? extends E> c)
  
   public ArrayList(int initialCapacity)
  

• So when we pass the Arrays.asList returned object, i.e., List(AbstractList) to the second constructor above, it will create a new dynamic array (this array size increases as we add more elements than its capacity and also the new elements will not affect the original array) shallow copying the original array (shallow copy means it copies over the references only and does not create a new set of same objects as in original array)

Answer 6

Note that, in Java 8, 'ia' above must be Integer[] and not int[]. Arrays.asList() of an int array returns a list with a single element. When using the OP's code snippet, the compiler will catch the issue, but some methods (e.g., Collections.shuffle()) will silently fail to do what you expect.

Answer 7

Summary of the difference -

When a list is created without using the new, the operator Arrays.asList() method returns a wrapper which means:

1. you can perform an add/update operation.

2. the changes done in the original array will be reflected to List as well and vice versa.

Answer 8

Well, this is because ArrayList resulting from Arrays.asList() is not of the type java.util.ArrayList.

Arrays.asList() creates an ArrayList of type java.util.Arrays$ArrayList which does not extend java.util.ArrayList, but only extends java.util.AbstractList.

Answer 9

List<Integer> list1 = new ArrayList<Integer>(Arrays.asList(ia));  //copy

In this case, list1 is of type ArrayList.

List<Integer> list2 = Arrays.asList(ia);

Here, the list is returned as a List view, meaning it has only the methods attached to that interface. Hence why some methods are not allowed on list2.

ArrayList<Integer> list1 = new ArrayList<Integer>(Arrays.asList(ia));

Here, you are creating a new ArrayList. You're simply passing it a value in the constructor. This is not an example of casting. In casting, it might look more like this:

ArrayList list1 = (ArrayList)Arrays.asList(ia);

Answer 10

In response to some comments asking questions about the behaviour of Arrays.asList() since Java 8:

    int[] arr1 = {1,2,3};
    /* 
       Arrays are objects in Java, internally int[] will be represented by 
       an Integer Array object which when printed on console shall output
       a pattern such as 
       [I@address for 1-dim int array,
       [[I@address for 2-dim int array, 
       [[F@address for 2-dim float array etc. 
   */
    System.out.println(Arrays.asList(arr1)); 

    /* 
       The line below results in Compile time error as Arrays.asList(int[] array)
       returns List<int[]>. The returned list contains only one element 
       and that is the int[] {1,2,3} 
    */
    // List<Integer> list1 = Arrays.asList(arr1);

    /* 
       Arrays.asList(arr1) is  Arrays$ArrayList object whose only element is int[] array
       so the line below prints [[I@...], where [I@... is the array object.
    */
    System.out.println(Arrays.asList(arr1)); 

    /* 
     This prints [I@..., the actual array object stored as single element 
     in the Arrays$ArrayList object. 
    */
    System.out.println(Arrays.asList(arr1).get(0));

    // prints the contents of array [1,2,3]
    System.out.println(Arrays.toString(Arrays.asList(arr1).get(0)));

    Integer[] arr2 = {1,2,3};
    /* 
     Arrays.asList(arr) is  Arrays$ArrayList object which is 
     a wrapper list object containing three elements 1,2,3.
     Technically, it is pointing to the original Integer[] array 
    */
    List<Integer> list2 = Arrays.asList(arr2);

    // prints the contents of list [1,2,3]
    System.out.println(list2);

Answer 11

package com.copy;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Iterator;
import java.util.List;

public class CopyArray {

    public static void main(String[] args) {
        List<Integer> list1, list2 = null;
        Integer[] intarr = { 3, 4, 2, 1 };
        list1 = new ArrayList<Integer>(Arrays.asList(intarr));
        list1.add(30);
        list2 = Arrays.asList(intarr);
        // list2.add(40); Here, we can't modify the existing list,because it's a wrapper
        System.out.println("List1");
        Iterator<Integer> itr1 = list1.iterator();
        while (itr1.hasNext()) {
            System.out.println(itr1.next());
        }
        System.out.println("List2");
        Iterator<Integer> itr2 = list2.iterator();
        while (itr2.hasNext()) {
            System.out.println(itr2.next());
        }
    }
}

Answer 12

1.List<Integer> list1 = new ArrayList<Integer>(Arrays.asList(ia));  //copy
2.List<Integer> list2 = Arrays.asList(ia);

In line 2, Arrays.asList(ia) returns a List reference of inner class object defined within Arrays, which is also called ArrayList but is private and only extends AbstractList. This means what returned from Arrays.asList(ia) is a class object different from what you get from new ArrayList<Integer>.

You cannot use some operations to line 2 because the inner private class within Arrays does not provide those methods.

Take a look at this link and see what you can do with the private inner class: http://grepcode.com/file/repository.grepcode.com/java/root/jdk/openjdk/6-b14/java/util/Arrays.java#Arrays.ArrayList

Line 1 creates a new ArrayList object copying elements from what you get from line 2. So you can do whatever you want since java.util.ArrayList provides all those methods.

Answer 13

1. First, let's see what this does:

   Arrays.asList(ia)
   

   It takes an array ia and creates a wrapper that implements List<Integer>, which makes the original array available as a list. Nothing is copied and all, only a single wrapper object is created. Operations on the list wrapper are propagated to the original array. This means that if you shuffle the list wrapper, the original array is shuffled as well, if you overwrite an element, it gets overwritten in the original array, etc. Of course, some List operations aren't allowed on the wrapper, like adding or removing elements from the list, you can only read or overwrite the elements.

   Note that the list wrapper doesn't extend ArrayList - it's a different kind of object. ArrayLists have their own, internal array, in which they store their elements, and are able to resize the internal arrays etc. The wrapper doesn't have its own internal array, it only propagates operations to the array given to it.

2. On the other hand, if you subsequently create a new array as

   new ArrayList<Integer>(Arrays.asList(ia))
   

   then you create new ArrayList, which is a full, independent copy of the original one. Although here you create the wrapper using Arrays.asList as well, it is used only during the construction of the new ArrayList and is garbage-collected afterwards. The structure of this new ArrayList is completely independent of the original array. It contains the same elements (both the original array and this new ArrayList reference the same integers in memory), but it creates a new, internal array, that holds the references. So when you shuffle it, add, remove elements etc., the original array is unchanged.