Initialization of an ArrayList in one line

0 votes

I wanted to create a list of options for testing purposes. At first, I did this:

ArrayList<String> places = new ArrayList<String>();
places.add("Buenos Aires");
places.add("Córdoba");
places.add("La Plata");

Then, I refactored the code as follows:

ArrayList<String> places = new ArrayList<String>(
    Arrays.asList("Buenos Aires", "Córdoba", "La Plata"));

Is there a better way to do this?

Dec 21, 2020 in Java by Rajiv
• 8,880 points
47 views

1 answer to this question.

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Actually, probably the "best" way to initialize the ArrayList is the method you wrote, as it does not need to create a new List in any way:

ArrayList<String> list = new ArrayList<String>();
list.add("A");
list.add("B");
list.add("C");

The catch is that there is quite a bit of typing required to refer to that list instance.

There are alternatives, such as making an anonymous inner class with an instance initializer (also known as an "double brace initialization"):

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

However, I'm not too fond of that method because what you end up with is a subclass of ArrayList which has an instance initializer, and that class is created just to create one object -- that just seems like a little bit overkill to me.

What would have been nice was if the Collection Literals proposal for Project Coin was accepted (it was slated to be introduced in Java 7, but it's not likely to be part of Java 8 either.):

List<String> list = ["A", "B", "C"];

Unfortunately it won't help you here, as it will initialize an immutable List rather than an ArrayList, and furthermore, it's not available yet, if it ever will be.

answered Dec 21, 2020 by Gitika
• 65,870 points

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