Initialization of an ArrayList in one line
I want 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?
arraylist
Collections
Java
- asked 9 years ago
- B Butts
2Answer
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 ArrayListwhich 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 8 years ago
- Sandy Hook
It would be simpler if you were to just declare it as a List - does it have to be an ArrayList?
List<String> places = Arrays.asList("Buenos Aires", "Córdoba", "La Plata");
Or if you have only one element:
List<String> places = Collections.singletonList("Buenos Aires");
This would mean that places is immutable (trying to change it will cause an exception to be thrown).
To make a mutable list that is a concrete ArrayList you can create an ArrayList from the immutable list:
ArrayList<String> places = new ArrayList<>(Arrays.asList("Buenos Aires", "Córdoba", "La Plata"));
- answered 8 years ago
- Sunny Solu
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