I have an array that is initialised like:

Element[] array = {new Element(1),new Element(2),new Element(3)};

I would like to convert this array into an object of the ArrayList class.

ArrayList<Element> arraylist = ???;

I am sure I have done this before, but the solution is sitting just at the edge of my memory.

share|improve this question
90  
+1 for being the first hit on google when searching for "java construct arraylist from array". Thanks! – Greg Hewgill Oct 8 '09 at 22:45

7 Answers

up vote 574 down vote accepted
new ArrayList<Element>(Arrays.asList(array))
share|improve this answer
41  
Yep. And in the (most common) case where you just want a list, the new ArrayList call is unecessary as well. – Calum Oct 1 '08 at 14:41
10  
@Luron - just use List<ClassName> list = Arrays.asList(array) – Pool Jun 29 '11 at 15:18
28  
@Calum and @Pool - as noted below in Alex Miller's answer, using Arrays.asList(array) without passing it into a new ArrayList object will fix the size of the list. One of the more common reasons to use an ArrayList is to be able to dynamically change its size, and your suggestion would prevent this. – Code Jockey Sep 26 '11 at 19:04
3  
I'd like to point out that the new ArrayList<Element>(...) part is necessary if you want to perform mutative (like remove) operations with an iterator. – markbaldy Nov 23 '12 at 5:28
6  
Arrays.asList() is a horrible function, and you should never just use its return value as is. It breaks the List template, so always use it in the form indicated here, even if it does seem redundant. Good answer. – Adam Jan 23 at 3:28
show 8 more comments

Given:

Element[] array = new Element[] { new Element(1),new Element(2),new Element(3) };

The simplest answer is to do:

List<Element> list = Arrays.asList(array);

This will work fine. But some caveats:

  1. The list returned from asList has fixed size. So, if you want to be able to add or remove elements from the returned list in your code, you'll need to wrap it in a new ArrayList. Otherwise you'll get an UnsupportedOperationException.
  2. The list returned from asList() is backed by the original array. If you modify the original array, the list will be modified as well. This may be surprising.
share|improve this answer
13  
+1 for the UnsupportedOperationException caveat. – Senthil Kumar May 1 '12 at 10:17
What is the time complexity of both operations? I mean with and without using explicit arraylist construction. – damned May 23 '12 at 2:17
Arrays.asList() merely creates an ArrayList by wrapping the existing array so it is O(1). – Alex Miller May 23 '12 at 13:15
Wrapping in a new ArrayList() will cause all elements of the fixed size list to be iterated and added to the new ArrayList so is O(n). – Alex Miller May 23 '12 at 13:21
the implementation of List returned by asList() does not implement several of List's methods (like add(), remove(), clear(), etc...) which explains the UnsupportedOperationException. definitely a caveat... – sethro Nov 14 '12 at 19:33

Since this question is pretty old, it suprises me that nobody suggested the simplest form yet:

List<Element> arraylist = Arrays.asList(new Element(1),new Element(2),new Element(3));

As of Java 5, Arrays.asList() takes a varargs parameter and you don't have to construct the array explicit.

share|improve this answer
2  
Thanks Tim; good point. I have up-voted this but will retain the other accepted answer as the question was specifically about the conversion of an array to a List. – Ron Tuffin Jun 22 '11 at 13:47
4  
Fair enough. Maybe that helps someone else. – Tim Büthe Jun 27 '11 at 8:00
new ArrayList<T>(Arrays.asList(myArray));
share|improve this answer
19  
you need to type faster ;) – Tom Oct 1 '08 at 14:41
7  
I was just thinking the same thing! :) – Bill the Lizard Oct 1 '08 at 14:42
1  
This one is actually more correct (Tom's is missing the type annotation) :) – Calum Oct 1 '08 at 14:42
you're right, I updated my answer – Tom Oct 1 '08 at 14:44
10  
and again you should type faster :) – Tom Oct 1 '08 at 14:46
show 1 more comment

(old thread, but just 2 cents as none mention Guava or other libs and some other details)

If You Can, Use Guava

It's worth pointing out the Guava way, which greatly simplifies these shenanigans:

Usage

For an Immutable List

Use the ImmutableList class and its of() and copyOf() factory methods (elements can't be null):

List<String> il = ImmutableList.of("string", "elements");  // from varargs
List<String> il = ImmutableList.copyOf(aStringArray);      // from array

For A Mutable List

Use the Lists class and its newArrayList() factory methods:

List<String> l1 = Lists.newArrayList(anotherListOrCollection);    // from collection
List<String> l2 = Lists.newArrayList(aStringArray);               // from array
List<String> l3 = Lists.newArrayList("or", "string", "elements"); // from varargs

Please also note the similar methods for other data structures in other classes, for instance in Sets.

Why Guava?

The main attraction could be to reduce the clutter due to generics for type-safety, as the use of the Guava factory methods allow the types to be inferred most of the time. However, this argument holds less water since Java 7 arrived with the new diamond operator.

But it's not the only reason (and Java 7 isn't everywhere yet): the shorthand syntax is also very handy, and the methods initializers, as seen above, allow to write more expressive code. You do in one Guava call what takes 2 with the current Java Collections.


If You Can't...

For an Immutable List

Use the JDK's Arrays class and its asList() factory method:

List<String> l1 = Arrays.asList(anArrayOfElements);
List<String> l2 = Arrays.asList("element1", "element2");

Note that the returned type for asList() says ArrayList, but it's not java.util.ArrayList. It's an inner type, which copies ArrayList but actually directly references the past array and forbis modifications. See the next step if you need a mutable list.

For a Mutable List

Same as above, but wrapped with an actual java.util.ArrayList:

List<String> l1  = new ArrayList<String>(Arrays.asList(array));    // Java 1.5 to 1.6
List<String> l1b = new ArrayList<>(Arrays.asList(array));          // Java 1.7+
List<String> l2  = new ArrayList<String>(Arrays.asList("a", "b")); // Java 1.5 to 1.6
List<String> l2b = new ArrayList<>(Arrays.asList("a", "b"));       // Java 1.7+

For Educational Purposes: The Good ol' Manual Way

// for Java 1.5+
static <T> List<T> arrayToList(final T[] array) {
  final List<String> l = new ArrayList<String>(array.length);

  for (final String s : array) {
    l.add(s);
  }
  return (l);
}

// for Java < 1.5 (no generics, no compile-time type-safety, boo!)
static List arrayToList(final Object[] array) {
  final List l = new ArrayList(array.length);

  for (int i = 0; i < array.length; i++) {
    l.add(array[i]);
  }
  return (l);
}
share|improve this answer
2  
+1 But note that the List returned by Arrays.asList is mutable in that you can still set elements - it just isn't resizable. For immutable lists without Guava you might mention Collections.unmodifiableList. – Paul Bellora Jan 10 at 2:54

Another way (although essentially equivalent to the new ArrayList(Arrays.asList(array)) solution performance-wise:

Collections.addAll(arraylist, array);
share|improve this answer
This way lets you have a mutable arraylist, though, thanks! – rogerdpack Dec 3 '12 at 20:38

You probably just need a List, not an ArrayList. In that case you can just do:

List<Element> arraylist = Arrays.asList(array);
share|improve this answer
1  
That will be backed by the original input array, which is why you (probably) want to wrap it in a new ArrayList. – Bill the Lizard Oct 1 '08 at 14:46
6  
Be careful with this solution. If you look, Arrays ISN'T returning a true java.util.ArrayList. It's returning an inner class that implements the required methods, but you cannot change the memebers in the list. It's merely a wrapper around an array. – Mikezx6r Oct 1 '08 at 14:47
You can cast the List<Element> item to an ArrayList<Element> – monksy Oct 9 '09 at 22:48
5  
@Mikezx6r: little correction: it's a fixed-size list. You can change the elements of the list (set method), you cannot change the size of the list (not add or remove elements)! – Carlos Heuberger Dec 4 '09 at 13:10

protected by Robert Harvey May 11 '11 at 3:26

This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site.

Not the answer you're looking for? Browse other questions tagged or ask your own question.