ArrayList initialization equivalent to array initialization

Question

I am aware that you can initialize an array during instantiation as follows:

String[] names = new String[] {"Ryan", "Julie", "Bob"};

Is there a way to do the same thing with an ArrayList? Or must I add the contents individually with array.add()?

Answer 1

Arrays.asList can help here:

new ArrayList<Integer>(Arrays.asList(1,2,3,5,8,13,21));

Answer 2

Yes.

new ArrayList<String>(){{
   add("A");
   add("B");
}}

What this is actually doing is creating a class derived from ArrayList<String> (the outer set of braces do this) and then declare a static initialiser (the inner set of braces). This is actually an inner class of the containing class, and so it'll have an implicit this pointer. Not a problem unless you want to serialise it, or you're expecting the outer class to be garbage collected.

I understand that Java 7 will provide additional language constructs to do precisely what you want.

EDIT: recent Java versions provide more usable functions for creating such collections, and are worth investigating over the above (provided at a time prior to these versions)

Answer 3

Here is the closest you can get:

ArrayList<String> list = new ArrayList(Arrays.asList("Ryan", "Julie", "Bob"));

You can go even simpler with:

List<String> list = Arrays.asList("Ryan", "Julie", "Bob")

Looking at the source for Arrays.asList, it constructs an ArrayList, but by default is cast to List. So you could do this (but not reliably for new JDKs):

ArrayList<String> list = (ArrayList<String>)Arrays.asList("Ryan", "Julie", "Bob")

Answer 4

Arrays.asList("Ryan", "Julie", "Bob");

Answer 5

Well, in Java there's no literal syntax for lists, so you have to do .add().

If you have a lot of elements, it's a bit verbose, but you could either:

1. use Groovy or something like that
2. use Arrays.asList(array)

2 would look something like:

String[] elements = new String[] {"Ryan", "Julie", "Bob"};
List list = new ArrayList(Arrays.asList(elements));

This results in some unnecessary object creation though.

Answer 6

The selected answer is: ArrayList<Integer>(Arrays.asList(1,2,3,5,8,13,21));

However, its important to understand the selected answer internally copies the elements several times before creating the final array, and that there is a way to reduce some of that redundancy.

Lets start by understanding what is going on:

1. First, the elements are copied into the Arrays.ArrayList<T> created by the static factory Arrays.asList(T...).

   This does not the produce the same class as java.lang.ArrayListdespite having the same simple class name. It does not implement methods like remove(int) despite having a List interface. If you call those methods it will throw an UnspportedMethodException. But if all you need is a fixed-sized list, you can stop here.

2. Next the Arrays.ArrayList<T> constructed in #1 gets passed to the constructor ArrayList<>(Collection<T>) where the collection.toArray() method is called to clone it.

   public ArrayList(Collection<? extends E> collection) {
   ......
   Object[] a = collection.toArray();
   }
   

3. Next the constructor decides whether to adopt the cloned array, or copy it again to remove the subclass type. Since Arrays.asList(T...) internally uses an array of type T, the very same one we passed as the parameter, the constructor always rejects using the clone unless T is a pure Object. (E.g. String, Integer, etc all get copied again, because they extend Object).

   if (a.getClass() != Object[].class) {      
       //Arrays.asList(T...) is always true here 
       //when T subclasses object
       Object[] newArray = new Object[a.length];
       System.arraycopy(a, 0, newArray, 0, a.length);
       a = newArray;
   }
   array = a;
   size = a.length;
   

Thus, our data was copied 3x just to explicitly initialize the ArrayList. We could get it down to 2x if we force Arrays.AsList(T...) to construct an Object[] array, so that ArrayList can later adopt it, which can be done as follows:

(List<Integer>)(List<?>) new ArrayList<>(Arrays.asList((Object) 1, 2 ,3, 4, 5));

Or maybe just adding the elements after creation might still be the most efficient.

Answer 7

How about this one.

ArrayList<String> names = new ArrayList<String>();
Collections.addAll(names, "Ryan", "Julie", "Bob");

Answer 8

This is how it is done using the fluent interface of the op4j Java library (1.1. was released Dec '10) :-

List<String> names = Op.onListFor("Ryan", "Julie", "Bob").get();

It's a very cool library that saves you a tonne of time.