How to initialize List<T> in Kotlin?

Question

I see Kotlin has a List<out E> collection and I was wondering about different ways to initialize one. In Java, I could write:

List<String> geeks = Arrays.asList("Fowler", "Beck", "Evans");

How can I achieve the same in Kotlin?

Answer 1

listOf顶级函数来救援：

val geeks = listOf("Fowler", "Beck", "Evans")

Answer 2

Both the upvoted answers by Ilya and gmariotti are good and correct. Some alternatives are however spread out in comments, and some are not mentioned at all.

This answer includes a summary of the already given ones, along with clarifications and a couple of other alternatives.

Immutable lists ( List )

Immutable, or read-only lists, are lists which cannot have elements added or removed.

• As Ilya points out, listOf() often does what you want. This creates an immutable list, similar to Arrays.asList in Java.
• As frogcoder states in a comment, emptyList() does the same, but naturally returns an empty list.
• listOfNotNull() returns an immutable list excluding all null elements.

Mutable lists ( MutableList )

Mutable lists can have elements added or removed.

• gmariotti suggests using mutableListOf() , which typically is what you want when you need to add or remove elements from the list.
• Greg T gives the alternative, arrayListOf() . This creates a mutable ArrayList . In case you really want an ArrayList implementation, use this over mutableListOf() .
• For other List implementations, which have not got any convenience functions, they can be initialized as, for example, val list = LinkedList<String>() . That is simply create the object by calling its constructor. Use this only if you really want, for example, a LinkedList implementation.

Answer 3

Just for adding more info, Kotlin offers both immutable List and MutableList that can be initialized with listOf and mutableListOf . If you're more interested in what Kotlin offers regarding Collections, you can go to the official reference docs at Collections .

Answer 4

Let me explain some use-cases : let's create an immutable(non changeable) list with initializing items :

val myList = listOf("one" , "two" , "three")

let's create a Mutable (changeable) list with initializing fields :

val myList = mutableListOf("one" , "two" , "three")

Let's declare an immutable(non changeable) and then instantiate it :

lateinit var myList : List<String>
// and then in the code :
myList = listOf("one" , "two" , "three")

And finally add some extra items to each :

val myList = listOf("one" , "two" , "three")
myList.add() //Unresolved reference : add, no add method here as it is non mutable 

val myMutableList = mutableListOf("one" , "two" , "three")
myMutableList.add("four") // it's ok 

Answer 5

这样就可以在Kotlin中初始化List

val alphabates : List<String> = listOf("a", "b", "c")

Answer 6

There is one more way to build a list in Kotlin that is as of this writing in the experimental state but hopefully should change soon.

inline fun <E> buildList(builderAction: MutableList<E>.() -> Unit): List<E>

Builds a new read-only List by populating a MutableList using the given builderAction and returning a read-only list with the same elements.

Example:

val list = buildList {
    testDataGenerator.fromJson("/src/test/resources/data.json").forEach {
        add(dao.insert(it))
    }
}

For further reading check the official doc .

Answer 7

如果你想在没有类型的情况下初始化：

var temp: ArrayList<String> = ArrayList()