六,Kotlin自定义高阶函数

1,自定义高阶函数接收一个函数类型的参数

定义高阶函数,
传入两个参数，
传入一个函数,
函数结果抽象,来实现他们不同的逻辑

private fun resultNumber(num1: Int, num2: Int, result: (Int, Int) -> Int): Int {
    return result(num1, num2)
}

调用函数

fun main() {
    val result1 = resultNumber(1, 2) { num1, num2 ->
        num1 + num2
    }
    println("result1 = $result1")
    val result2 = resultNumber(1, 2) { num1, num2 ->
        num1 * num2
    }
    println("result2 = $result2")
    val result3 = resultNumber(1, 2) { num1, num2 ->
        num1 % num2
    }
    println("result3 = $result3")
}

result1 = 3
result2 = 2
result3 = 1

2,自定义高阶函数将函数类型作为返回值

//定义一个类

class UserEntity(var userId: Long, var userName: String, var age: Int = 0) {
    constructor() : this(0, "", 0)
    override fun toString(): String {
        return "userName:$userName,userId:$userId,age:$age"
    }
}

定义函数

//此方法用户判读用户是否为成年人.
fun isAdult(): (UserEntity) -> Boolean {
    return { entity -> entity.age >= 18 }
}

使用

fun main() {
    val user1 = UserEntity(100, "华晨宇", 30)
    val user2 = UserEntity(101, "张碧晨", 31)
    val user3 = UserEntity(103, "哇哈哈", 10)
    val isAdultUser = isAdult()
    println("${user1.userName}是 ${if (isAdultUser(user1)) "成年" else "儿童"}")
    println("${user2.userName}是 ${if (isAdultUser(user2)) "成年" else "儿童"}")
    println("${user3.userName}是 ${if (isAdult()(user3)) "成年" else "儿童"}")
}

华晨宇是 成年
张碧晨是 成年
哇哈哈是 儿童