Kotlin:val mutableList vs var immutableList.什么时候用哪个？

Question

我们鼓励尽可能多地使用不可变变量.但是,如果我有时需要修改列表,我想知道我应该使用哪种方法......

1. val mutableList = mutableListOf()在那里我可以只add,remove因此

要么

2. var immutableList = listOf()我+每次进行更改时都会创建一个新列表(使用过滤器或).

我猜有一种不同的情景,一种比另一种更受欢迎.因此想知道何时应该使用其他等.

Answer 1

可变和不可变列表增加了模型的设计清晰度.
这是为了迫使开发人员思考和澄清收集的目的.

1. 如果集合将作为设计的一部分进行更改,请使用可变集合
2. 如果model仅用于查看,请使用不可变列表

目的val和var不同于不可变和可变列表.
val和var关键字谈论如何处理变量的值/引用.

• var - 可以在任何时间点更改分配给变量的值/引用.
• val - 值/引用只能为变量赋值一次,并且在执行后的某个时间点不能更改.

在Kotlin中将可变列表分配给val并向其添加元素是完全有效的.

val a = mutableListOf(1,2,3,4)
a.add(5)
println(a)

将输出

[1, 2, 3, 4, 5]

Answer 2

val -> 你可能认为你不能为变量重新赋值。

//that is ok
var a:Int = 1
a=2
//Even you can reassign but you can't change its type
a= "string"  //that's wrong

//that is wrong
val b:Int = 1
b = 2

ListOf -> 你可能认为你不能插入/删除/更改列表中的任何元素（不能对列表的内容做任何事情）

var list:List<Int> = listOf(1,2,3,4) //[1,2,3,4]
//you can read list
list.get(0)
list[0]
//but you can't change(/write) the content of the list (insert/delete/alter)
list.set(0, 100)
list.add(5)
list.removeAt(0)

var mutableList:MutableList<Int> = mutableListOf(1,2,3,4) //[1,2,3,4]
//you can read and write
mutableList.get(0)
mutableList.set(0, 100) //[100,2,3,4]
mutableList.add(5)      //[100,2,3,4,5]
mutableList.removeAt(0) //[2,3,4,5]

所以结合他们两个，你会得到四个案例

情况 1：var mutableList:MutableList = mutableListOf(1,2,3,4)

//you can reassign 
mutableList = mutableListOf(4,5,6,7) //[4,5,6,7]
//you can alter the content 
mutableList.set(0, 100) //[100,5,6,7]
mutableList.add(8)      //[100,5,6,7,8]
mutableList.removeAt(0) //[5,6,7,8]

情况 2：val mutableList:MutableList = mutableListOf(1,2,3,4)

//you can't reassign 
mutableList = mutableListOf(4,5,6,7) //that's wrong

//you can alter the content 
mutableList.set(0, 100) //[100,2,3,4]
mutableList.add(8)      //[100,2,3,4,8]
mutableList.removeAt(0) //[2,3,4,8]

情况 3：var list:List = ListOf(1,2,3,4)

//you can reassign 
list= ListOf(4,5,6,7) //[4,5,6,7]

//you can't alter the content 
list.set(0, 100) //that's wrong
list.add(8)      //that's wrong
list.removeAt(0) //that's wrong

情况 4：val list:List = ListOf(1,2,3,4)

//you can't reassign 
list= ListOf(4,5,6,7) //that's wrong

//you can't alter the content 
list.set(0, 100) //that's wrong
list.add(8)      //that's wrong
list.removeAt(0) //that's wrong

//the only thing you can do is Read
list.get(0)  //return 1
list[0]      //return 1