使用泛型参数运行函数而不使用asInstanceOf

Question

这是我的示例scala代码：

object App {
  abstract class BaseAction
  type ApiAction[T <: BaseAction] = (T) => Unit

  case class FirstAction(name: String) extends BaseAction
  case class SecondAction(surname: String) extends BaseAction

  def action1[Z <: BaseAction] = {  
    (a: Z) => { // Here i'would like to have a: FirstAction
      val z = a.asInstanceOf[FirstAction]
      println("Running action: " + z.name )
    }
   }

 def action2[Z <: BaseAction] = {
  (a: Z) => { // Here i'would like to have a: SecondAction
  val z = a.asInstanceOf[SecondAction]
  println("Running action " + z.surname )
 }
}

  def myActions[T <: BaseAction] = Map[String, ApiAction[T]]("a1" -> action1[T], "a2" -> action2[T])

  myActions("a1")(FirstAction("Action 1"))
  myActions("a2")(SecondAction("Action 2"))
}

我有很少的动作功能，它们做了不同的事情。 每个动作函数都接收一个参数：action class，其中所有动作类都继承自BaseAction抽象类。

函数myActions是actionName到action函数的Map。

我的代码正在运行，但我认为使用asInstanceOf并不是一个好习惯，我想知道如何只使用泛型类型编写此代码，而不使用asInstanceOf。

Answer 1

问题是你做了很多“你不应该做这些事情”。

我将尝试给你“更好”（按照大多数scala人）的方式来写同样的东西。 这些更改包括使用来自类型边界的信息以及自定义类型ApiAction来编写更可预测和有组织的代码。

首先你有以下抽象，

  abstract class BaseAction

  type ApiAction[T <: BaseAction] = (T) => Unit

  case class FirstAction(name: String) extends BaseAction
  case class SecondAction(surname: String) extends BaseAction

现在您可以使用这些抽象来编写您的Actions对象，

object MyActions {

  val action1: ApiAction[FirstAction] = {
    case FirstAction(name) => println("Running action :: " + name)
  }

  val action1Other: ApiAction[FirstAction] = (fa: FirstAction) => {
    println("Running action :: " + fa.name)
  }

  val action2: ApiAction[SecondAction] = {
    case SecondAction(surname) => println("Running action :: " + surname)
  }

  val action2Other: ApiAction[SecondAction] = (sa: SecondAction) => {
    println("Running action :: " + sa.surname)
  }

  // but lets say you wanted a generic ApiAction
  val actionGeneric: ApiAction[BaseAction] = {
    case FirstAction(name) => println("Running action :: " + name)
    case SecondAction(surname) => println("Running action :: " + surname)
  }

}

现在，您可以在应用中使用这些“操作”，

object MyApp extends App {

  MyActions.action1(FirstAction("Action 1"))

  MyActions.action1Other(FirstAction("Action 1 Other"))

  MyActions.actionGeneric(FirstAction("Action 1 Generic"))

  MyActions.action2(SecondAction("Action 2"))

  MyActions.action2Other(SecondAction("Action 2 Other"))

  MyActions.actionGeneric(SecondAction("Action 2 Generic"))
}

Answer 2

您可以使用惯用的Scala模式匹配而不是asInstanceOf ：

val z = a match {
  case FirstAction(name) => println("Running action " + name)
  case _ => println("Error")
}

请注意，模式匹配仍然使用isInstanceof + asInstanceOf ，但它被认为是一种很好的做法，不像直接调用asInstanceOf 。

顺便说一下，组织你的代码可能是明智的，这样你只匹配一次，而不是有两个单独的“第一个动作或错误”和“第二个动作或错误”块：

def action[Z <: BaseAction] = {
  (a: Z) => a match {
    case FirstAction(name) => println("Running action " + name)
    case SecondAction(surname) => println("Running action " + surname)
    case _ => println("Error")
  }
}

def myActions[T <: BaseAction] = Map[String, ApiAction[T]]("a1" -> action[T], "a2" -> action[T])

myActions("a1")(FirstAction("Action 1"))
myActions("a2")(SecondAction("Action 2"))

// output:
// Running action Action 1
// Running action Action 2