Initializing ArrayList with some predefined values

Question

I have an sample program as shown.

I want my ArrayList symbolsPresent to be initialized with some predefined symbols: ONE, TWO, THREE, and FOUR.

symbolsPresent.add("ONE");
symbolsPresent.add("TWO");
symbolsPresent.add("THREE");
symbolsPresent.add("FOUR");

import java.util.ArrayList;

public class Test {

    private ArrayList<String> symbolsPresent = new ArrayList<String>();

    public ArrayList<String> getSymbolsPresent() {
        return symbolsPresent;
    }

    public void setSymbolsPresent(ArrayList<String> symbolsPresent) {
        this.symbolsPresent = symbolsPresent;
    }

    public static void main(String args[]) {    
        Test t = new Test();
        System.out.println("Symbols Present is" + t.symbolsPresent);

    }    
}

Is that possible?

Will this help? http://stackoverflow.com/questions/1005073/initialization-of-an-arraylist-in-one-line — Jops, Apr 24 '13 at 14:43

score 167 · Answer 1 · edited Dec 28 '15 at 09:50

167

try this

new String[] {"One","Two","Three","Four"};

or

List<String> places = Arrays.asList("One", "Two", "Three");

ARRAYS

edited Dec 28 '15 at 09:50

Boris

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answered Apr 24 '13 at 14:43

samba

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2

good solution, asList(...) returns new ArrayList<>(a) like the question's author requested – cahen Apr 24 '13 at 14:47
4

@CarlosHenriqueRodriguez asList(...) returns List not ArrayList – PermGenError Apr 24 '13 at 14:55
@PermGenError List is the return type, the method returns ArrayList. It's not java.util.ArrayList, it's an internal class with the same behavior – cahen Apr 24 '13 at 14:59
2

The list returned by `Arrays.asList(T...)` is capped in size to the size of the array supplied in the constructor. That may or may not be an issue, depending on usage. – Dilum Ranatunga Apr 24 '13 at 14:59

Reimeus · Answer 2 · 2014-02-05T21:35:07.213

59

Double brace initialization is an option:

List<String> symbolsPresent = new ArrayList<String>() {{
   add("ONE");
   add("TWO");
   add("THREE");
   add("FOUR");
}};

Note that the String generic type argument is necessary in the assigned expression as indicated by JLS §15.9

It is a compile-time error if a class instance creation expression declares an anonymous class using the "<>" form for the class's type arguments.

edited Feb 05 '14 at 21:35

answered Apr 24 '13 at 14:48

Reimeus

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interesting anonymous innterclass! – Ryan Apr 24 '13 at 14:52
If you add an object that was defined outside of this structure, you'll need to declare it `final` in order for it to compile. Read [this answer](http://stackoverflow.com/questions/4732544/why-are-only-final-variables-accessible-in-anonymous-class) if you want to know why. – May 22 '14 at 13:42
There's also a [performance penalty](http://stackoverflow.com/questions/924285/efficiency-of-java-double-brace-initialization), although it's significance is debatable. – May 22 '14 at 13:50
this is cool, but in Java 7 you can avoid specifying the type in the initialization: here http://docs.oracle.com/javase/7/docs/technotes/guides/language/type-inference-generic-instance-creation.html – Rose Jul 07 '15 at 21:14
@Rose that link only details standard `List` initialization, i.e. no anonymous inner class with values being added :) – Reimeus Jul 07 '15 at 21:43
@Reimeus my point is that instead of List symbolsPresent = new ArrayList() ..... you can do List symbolsPresent = new ArrayList<>() – Rose Jul 07 '15 at 23:49
that is an empty `List` which is not what the OP wanted... – Reimeus Jul 08 '15 at 01:15

score 21 · Answer 3 · answered Apr 24 '13 at 14:42

21

How about using overloaded ArrayList constructor.

 private ArrayList<String> symbolsPresent = new ArrayList<String>(Arrays.asList(new String[] {"One","Two","Three","Four"}));

answered Apr 24 '13 at 14:42

PermGenError

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5

wouldn't Arrays.asList() alone do the job? – cahen Apr 24 '13 at 14:49
4

@CarlosHenriqueRodriguez it would, but then you will have to have List on the left side. :) – PermGenError Apr 24 '13 at 14:54
Note that you can use the var-args syntax with Java 5 and later. – Dilum Ranatunga Apr 24 '13 at 15:00
2

If I am not mistaken, if you use Arrays.asList() and use directly the returned List, it would be fixed size. Basically it would still behave like an array, just wrapped in List interface. – Lukáš Řádek Jan 25 '17 at 00:40

score 13 · Answer 4 · answered Apr 24 '13 at 14:58

You can also use the varargs syntax to make your code cleaner:

Use the overloaded constructor:

ArrayList<String> list = new ArrayList<String>(Arrays.asList("a", "b", "c"));

Subclass ArrayList in a utils module:

public class MyArrayList<T> extends ArrayList<T> {
    public MyArrayList(T... values) {
        super(Arrays.asList(values));
    }
}

ArrayList<String> list = new MyArrayList<String>("a", "b", "c");

Or have a static factory method (my preferred approach):

public class Utils {
  public static <T> ArrayList<T> asArrayList(T... values) {
    return new ArrayList<T>(Arrays.asList(values));
  }
}

ArrayList<String> list = Utils.asArrayList("a", "b", "c");

score 10 · Answer 5 · edited Mar 22 '18 at 22:39

10

import com.google.common.collect.Lists;

...


ArrayList<String> getSymbolsPresent = Lists.newArrayList("item 1", "item 2");

...

edited Mar 22 '18 at 22:39

AngeloS

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answered Sep 10 '15 at 06:03

Kristy Hughes

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score 9 · Answer 6 · answered Nov 15 '15 at 11:40

9

You can use Java 8 Stream API.
You can create a Stream of objects and collect them as a List.

private List<String> symbolsPresent = Stream.of("ONE", "TWO", "THREE", "FOUR")
    .collect(Collectors.toList());

answered Nov 15 '15 at 11:40

Georgios Syngouroglou

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Voted up because streams provide a uniform mechanism for addressing a wide range of problems whereas the custom functions simply chew up mortal memory. – Mark Mullin Dec 24 '15 at 02:25

score 7 · Answer 7 · answered Nov 24 '16 at 22:47

7

public static final List<String> permissions = new ArrayList<String>() {{
    add("public_profile");
    add("email");
    add("user_birthday");
    add("user_about_me");
    add("user_location");
    add("user_likes");
    add("user_posts");
}};

answered Nov 24 '16 at 22:47

Narendra Sorathiya

3,770
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score 4 · Answer 8 · answered Apr 29 '18 at 23:33

Java 9 allows you to create an unmodifiable list with a single line of code using List.of factory:

public class Main {
    public static void main(String[] args) {
        List<String> examples = List.of("one", "two", "three");
        System.out.println(examples);
    }
}

Output:

[one, two, three]

mki · Answer 9 · 2013-04-24T18:43:14.707

Personnaly I like to do all the initialisations in the constructor

public Test()
{
  symbolsPresent = new ArrayList<String>();
  symbolsPresent.add("ONE");
  symbolsPresent.add("TWO");
  symbolsPresent.add("THREE");
  symbolsPresent.add("FOUR");
}

Edit : It is a choice of course and others prefer to initialize in the declaration. Both are valid, I have choosen the constructor because all type of initialitions are possible there (if you need a loop or parameters, ...). However I initialize the constants in the declaration on the top on the source.
The most important is to follow a rule that you like and be consistent in our classes.

score 2 · Answer 10 · answered Apr 24 '13 at 14:52

2

Also, if you want to enforce the List to be read-only (throws a UnsupportedOperationException if modified):

List<String> places = Collections.unmodifiableList(Arrays.asList("One", "Two", "Three"));

answered Apr 24 '13 at 14:52

ZachF

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score 2 · Answer 11 · answered Dec 07 '14 at 05:13

2

I would suggest to use Arrays.asList() for single line initialization. For different ways of declaring and initializing a List you can also refer Initialization of ArrayList in Java

answered Dec 07 '14 at 05:13

Manoj Kumar

111
3

Domenico Vacchiano · Answer 12 · 2015-06-09T14:53:57.263

1

I use a generic class that inherit from ArrayList and implement a constructor with a parameter with variable number or arguments :

public class MyArrayList<T> extends ArrayList<T> {
    public MyArrayList(T...items){
        for (T item : items) {
            this.add(item);
        }
    }
}

Example:

MyArrayList<String>myArrayList=new MyArrayList<String>("s1","s2","s2");

edited Jun 09 '15 at 14:53

answered Jun 09 '15 at 14:43

Domenico Vacchiano

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score 0 · Answer 13 · answered Dec 28 '15 at 10:37

If you just want to initialize outside of any method then use the initializer blocks :

import java.util.ArrayList;

public class Test {
private ArrayList<String> symbolsPresent = new ArrayList<String>();

// All you need is this block.
{
symbolsPresent = new ArrayList<String>();
symbolsPresent.add("ONE");
symbolsPresent.add("TWO");
symbolsPresent.add("THREE");
symbolsPresent.add("FOUR");
}


public ArrayList<String> getSymbolsPresent() {
    return symbolsPresent;
}

public void setSymbolsPresent(ArrayList<String> symbolsPresent) {
    this.symbolsPresent = symbolsPresent;
}

public static void main(String args[]) {    
    Test t = new Test();
    System.out.println("Symbols Present is" + t.symbolsPresent);

}    
}

Initializing ArrayList with some predefined values

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