Scala 高阶函数的Case类方法

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

我要做的是将公共位分解成f3，然后将f1和f2之间不同的位作为传递到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")
}

---

我所困惑的是，我如何才能在case类中传递一个方法函数？还有什么方法可以优雅地执行此操作吗？

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）

然后，您可以定义一些有用的函数，并使用currying使它们更易于以后使用：

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"))

---

在

f3

中，您可以简单地接受类型为

Action=>Action

的函数（我将使用

Action

而不是

Action

，以减少混淆）

然后，您可以定义一些有用的函数，并使用currying使它们更易于以后使用：

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"))