Java quick array list

Question

I have a very simple question:

How can I make this code more simple on Java:

ArrayList<String> s = new ArrayList<String>();
s.add("str1");
s.add("str hello");
s.add("str bye");
//...

Something like that:

ArrayList<String> s = {"a1", "str", "mystr"};

or that:

ArrayList<String> s = new ArrayList<String>("a1", "str", "mystr");

or that:

ArrayList<String> s = new ArrayList<String>();
s.addAll("a1", "str", "mystr");

or that:

ArrayList<String> s = new ArrayList<String>();
s.addAll(new ArrayElements("a1", "str", "mystr"));

I just want syntax hint. Thanks.

http://stackoverflow.com/questions/2760995/java-arraylist-initialization — Vlad, Oct 14 '11 at 12:22

score 41 · Answer 1 · answered Oct 14 '11 at 12:21

41

List<String> s = Arrays.asList("a1", "str", "mystr");

answered Oct 14 '11 at 12:21

dty

8

This returns a fixed-size list not a java.util.ArrayList. So it is good when no more adding or removing of elements is needed. – x22 Oct 14 '11 at 13:03
3

This is true. It wasn't clear from the OP whether this would be an issue or not. – dty Oct 14 '11 at 13:49

score 25 · Accepted Answer · answered Oct 14 '11 at 12:19

25

How about:

ArrayList<String> s = new ArrayList<String>();
Collections.addAll(s, "a1", "str", "mystr");

answered Oct 14 '11 at 12:19

NPE

score 5 · Answer 3 · answered Oct 14 '11 at 12:19

5

List<String> s = Arrays.asList(new String[] {"a1", "str", "mystr"});

answered Oct 14 '11 at 12:19

solendil

@Rob: the OP asks for simple, not for fast – amit Oct 14 '11 at 12:20
9

`List s = Arrays.asList("a1", "str", "mystr");` is even better – Eng.Fouad Oct 14 '11 at 12:22

score 5 · Answer 4 · answered Oct 14 '11 at 12:20

5

I would use Guava and its wonderful Lists class:

List<String> list = Lists.newArrayList("a1", "str", "mystr");

answered Oct 14 '11 at 12:20

Jon Skeet

score 5 · Answer 5 · answered Oct 14 '11 at 12:25

5

You can use double brace:

ArrayList<String> s = new ArrayList<String>()
{{
    add("str1");
    add("str hello");
    add("str bye");
    //...
}};

answered Oct 14 '11 at 12:25

Eng.Fouad

2

Rather than just producing esoteric syntax, wouldn't it be a good idea to explain to the OP what that is and how it works? – dty Oct 14 '11 at 12:26
1

dty, it is double brace initialization, it explained in http://c2.com/cgi/wiki?DoubleBraceInitialization in detail. – SerCe Sep 03 '14 at 17:03
There are reasons why you might not want to do this. This essentially creates a new anonymous inner class definition every time it's run, which can have side effects on a long running app, or if you execute this snippet frequently. It's often better to not use this form of initialization, because even if you understand the implications, someone on your team who may copy you may not. There are safer ways to do this – Julian May 29 '18 at 17:51

score 2 · Answer 6 · answered Jun 24 '19 at 20:07

Before or after Java 8 you can write:

ArrayList<String> s = 
    new ArrayList<>(Arrays.asList("str1", "str hello", "str bye"));

Since Java 8 you can write:

ArrayList<String> s =
    Stream.of("str1", "str hello", "str bye")
        .collect(Collectors.toCollection(ArrayList::new));

With Eclipse Collections you can write the following:

ArrayList<String> s =
    Lists.mutable.with("str1", "str hello", "str bye")
        .into(new ArrayList<>());

If you can use List instead of ArrayList:

List<String> s = Lists.mutable.with("str1", "str hello", "str bye");

Note: I am a committer for Eclipse Collections.

score 0 · Answer 7 · answered Sep 29 '17 at 17:07

If you're using Java 9, you could use List.of() :

List<String> s = List.of("str1", "str hello", "str bye");

But this would be an immutable list. If you need a mutable Arraylist:

ArrayList<String> s = new ArrayList<>(List.of("str1", "str hello", "str bye"))

7 Answers7