Scala深入浅出实战经典：23,Scala高阶函数实战详解

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package com.dt.scalaInAction.demo_023/** * Scala高阶函数实战详解 */object HighOrderFunctions {    def main(args: Array[String]): Unit = {        //1 打印阶梯状的*号//        1.to(9).map("*" * _).foreach(println)              //2 过滤1~9中的偶数//        1.to(9).filter(_ % 2 == 0).foreach(println _)  //row result: 2 4 6 8                //3 9*8*7*6*...*1//        println((1 to 9).reduceLeft(_ * _))                //4 按单词的长度升序排列        /*        "Spark is the most exciting thing happening in big data today"             .split(" ")             .sortWith(_.length < _.length)             .foreach(println)        */              //5        /*        val fun = ceil _        val num = 3.14        fun(num)        Array(3.14, 1.42, 2.0).map(fun)        */              //6//        val triple = (x: Double) => 3 * x//        Array(3.14, 1.42, 2.0).map((x: Double) => 3 * x).foreach(println)    //row result: 9.42 4.26 6.0//        Array(3.14, 1.42, 2.0).map{  (x: Double) => 3 * x }.foreach(println) //row result: 9.42 4.26 6.0                //7 高阶函数标准的定义方式   f: (Double) => Double   函数作为的参数类型定义  (参数值类型...) => 返回值类型        def high_order_functions(f: (Double) => Double) = f(0.25)//        println(high_order_functions(ceil _))//        println(high_order_functions(sqrt _))              //8   (x: Double)才是正真执行所要传入的参数        def mulBy(factor: Double) = (x: Double) => factor * x        val quintuple = mulBy(5)        println(quintuple(20))   //执行时 x=20     row result: 100.0                        //9 高阶函数的简化写法推导        println(high_order_functions((x: Double) => 3 * x))   //row result: 0.75        high_order_functions((x) => 3 * x )        high_order_functions(x => 3 * x )        high_order_functions(3 * _)                val fun2 = 3 * (_: Double)        val funn3:(Double) => Double = 3 * _   //_为返回类型(Double) => Double中的参数类型    }}

以上内容是从王家林老师DT大数据课程第23讲的学习笔记和个人整理。

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