在Java中,数组可以这样初始化:

int numbers[] = new int[] {10, 20, 30, 40, 50}

Kotlin的数组初始化是怎样的?


当前回答

我的回答补充了@maroun,这些是初始化数组的一些方法:

使用数组

val numbers = arrayOf(1,2,3,4,5)

使用严格的数组

val numbers = intArrayOf(1,2,3,4,5)

混合矩阵类型

val numbers = arrayOf(1,2,3.0,4f)

嵌套数组

val numbersInitials = intArrayOf(1,2,3,4,5)
val numbers = arrayOf(numbersInitials, arrayOf(6,7,8,9,10))

能够从动态代码开始

val numbers = Array(5){ it*2}

其他回答

这里有一个简单的例子

val id_1: Int = 1
val ids: IntArray = intArrayOf(id_1)

你可以试试这个:

var a = Array<Int>(5){0}

简单的方法:

整数:

var number = arrayOf< Int> (10,20,30,40,50)

保持所有数据类型

var number = arrayOf(10, "string value", 10.5)

您可以使用这些方法

var numbers=Array<Int>(size,init)
var numbers=IntArray(size,init)
var numbers= intArrayOf(1,2,3)

例子

var numbers = Array<Int>(5, { i -> 0 })

Init表示默认值(initialize)

I think one thing that is worth mentioning and isn't intuitive enough from the documentation is that, when you use a factory function to create an array and you specify it's size, the array is initialized with values that are equal to their index values. For example, in an array such as this: val array = Array(5, { i -> i }), the initial values assigned are [0,1,2,3,4] and not say, [0,0,0,0,0]. That is why from the documentation, val asc = Array(5, { i -> (i * i).toString() }) produces an answer of ["0", "1", "4", "9", "16"]