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doc: update docs/kotlin.md (#331)
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1250422131 committed Mar 16, 2023
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Expand Up @@ -606,6 +606,75 @@ fun main() {
```
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### 简单的高阶函数
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```kotlin
//注意啦,这里的num1AndNum2有个operation,它是接收了一个函数作为形参
fun num1AndNum2(num1: Int, num2: Int, operation: (Int, Int) -> Int): Int {
//让我们试着向operation传入参数
return operation(num1, num2)
}

fun plus(num1: Int, num2: Int): Int {
return num1 + num2
}

fun main(args: Array<String>) {
val total = num1AndNum2(100, 200, ::plus)
println(total)//300
//怎么样?我们利用传入一个函数来充当另一个函数的参数
}

```

还记得我们怎么在Java中用接口吗?

试着用函数参数简化它

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### 以匿名函数作为参数的函数

```kotlin
//operation是一个函数类型的参数哦
fun num1AndNum2(num1: Int, num2: Int, operation: (Int, Int) -> Int): Int {
return operation(num1, num2)
}


fun main(args: Array<String>) {
//这里我们定义一个匿名函数
val operation: (Int, Int) -> Int = { i: Int, i2: Int ->
i + i2
}
val total = num1AndNum2(100, 200, operation)
println(total) //300
}

```
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### Lambda表达式方式传入函数参数
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```kotlin
//我们还是不改变什么
fun num1AndNum2(num1: Int, num2: Int, operation: (Int, Int) -> Int): Int {
return operation(num1, num2)
}

fun main(args: Array<String>) {
//wow哦天哪,Lambda可以做到这样简洁
val total = num1AndNum2(100, 200) { n1, n2 ->
n1 + n2
}
println(total)
}
```

这里之所以可以把lambda写在外部,是因为operation是最后一个参数。
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