如何将Kotlin的MutableList初始化为空的MutableList? [英] How do I initialize Kotlin's MutableList to empty MutableList?

问题描述

似乎很简单，但是，如何将Kotlin的MutableList初始化为空的MutableList?

Seems so simple, but, how do I initialize Kotlin's MutableList to empty MutableList?

我可以用这种方式来破解它，但是我敢肯定有一些更容易的方法:

I could hack it this way, but I'm sure there is something easier available:

var pusta: List<Kolory> = emptyList()
var cos: MutableList<Kolory> = pusta.toArrayList()

推荐答案

您可以简单地编写:

val mutableList = mutableListOf<Kolory>()

这是最惯用的方式.

替代方法是

val mutableList : MutableList<Kolory> = arrayListOf()

或

val mutableList : MutableList<Kolory> = ArrayList()

这利用了这样的事实:像ArrayList这样的java类型通过编译器技巧隐式地实现了MutableList类型.

This is exploiting the fact that java types like ArrayList are implicitly implementing the type MutableList via a compiler trick.

这篇关于如何将Kotlin的MutableList初始化为空的MutableList?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！