具有案例类方法的高阶函数答案

【问题标题】：Higher Order Function with Case Class Method具有案例类方法的高阶函数
【发布时间】：2020-11-19 07:14:51
【问题描述】：

我正在尝试删除代码中的冗余。我想我可以用高阶函数来做到这一点。

我要做的是将公共位分解到f3中，然后将f1和f2之间不同的位作为传递给f3的参数。

object Example extends App {

  case class action(name: String, age: Int) {
    def setName(new_name: String): action = this.copy(name = new_name)
    def setAge(new_age: Int): action = this.copy(age = new_age)
  }

  def f1(x: action, increment: Int) = {
    // big block of code which does a....

    // a single line in the block calling
    val older_person = x setAge (x.age + increment)

    // big block of code which does b....
  }

  def f2(x: action, new_name: String) = {
    // big block of code which does a....

    // a single line in the block calling
    val new_name_person = x setName new_name

    // big block of code which does b....
  }

  /* Now as there is clearly a redundancy, which can be solved by higher order functions.
    I want to combine f1 and f2 into a single function. The function will take in the action, the value, and the 
    function to apply. It will then call the relevant function inside the method.
   */
  def f3[T](x: action)(f: T => action)(value: T) = {
   // big block of code which does a....
   // call x.f(value)
    val new_action = ??? 
   // big block of code which does b....
  }

  // then in my code I can call like this:
  // f3(x)(setAge)(100)
  // f3(x)(setName("new_name")
}

我很困惑的是如何传入一个作为案例类中的方法的函数？有没有一种优雅的方式来做到这一点？

【问题讨论】：

标签： scala higher-order-functions abstraction case-class

【解决方案1】：

在f3 中，您可以简单地接受Action => Action 类型的函数（我将使用Action 而不是action 以减少混淆）。

def f3(x: Action)(copy: Action => Action) = {
  // big block of code which does a....

  // a single line in the block calling
  val new_name_person = copy(x)

  // big block of code which does b....
}

然后你可以定义一些有用的函数，并使用currying让它们以后更容易使用：

object Action {
  def setName(name: String)(action: Action) = action.copy(name=name)
  def incAge(inc: Int)(action: Action) = action.copy(age=action.age+inc)
}

然后像这样使用它：

val x = Action("Foo", 42)
f3(x)(Action.incAge(100))
f3(x)(Action.setName("new_name"))

Try it in Scastie

【讨论】：