Scala:根据参数返回不同的类型 - 无需强制转换

时间:2015-03-29 14:53:13

标签: scala

我正在尝试编写一个函数,根据传入的参数,它可以返回不同的类型。要使用一个非常简单的例子,这是我的目标: 给出如下对象:

case object NameToken
case object SalaryToken
case object IsDirectorToken

val john: Employee

然后:

  • john.get(NameToken)将返回“John”(字符串),
  • john.get(SalaryToken)将返回50000(一个Int),
  • john.get(IsDirectorToken)将返回false(布尔值)。

以下是我想到的几种可能的实现方式。不言而喻,两者都很可怕。

可能的实施1:

trait Token[A]
case object NameToken extends Token[String]
case object SalaryToken extends Token[Int]
case object IsDirectorToken extends Token[Boolean]

case class Employee(name: String, salary: Int, isDirector: Boolean) {
  def get[A](t: Token[A]): A = t match {
    case NameToken => name.asInstanceOf[A]
    case SalaryToken => salary.asInstanceOf[A]
    case IsDirectorToken => isDirector.asInstanceOf[A]
  }
}

可能的实施2:

trait Token2 {
  type returnType
}
case object NameToken2 extends Token2 {
  type returnType = String
}
case object SalaryToken2 extends Token2 {
  type returnType = Int
}
case object IsDirectorToken2 extends Token2  {
  type returnType = Boolean
}

case class Employee2(name: String, salary: Int, isDirector: Boolean) {
  def get(t: Token2): t.returnType = t match {
    case NameToken2 => name.asInstanceOf[t.returnType]
    case SalaryToken2 => salary.asInstanceOf[t.returnType]
    case IsDirectorToken2 => isDirector.asInstanceOf[t.returnType]
  }
}
然而,他们两个都只是对那个演员来说很糟糕。

我可以更聪明地解决这个问题吗?

谢谢。

2 个答案:

答案 0 :(得分:4)

class Employee(val name: String, val salary: Int, val isDirector: Boolean) {
  def get[T](t: Token[T]): T = t.value(this)
}

trait Token[T] { def value(e: Employee): T }
object NameToken extends Token[String] { def value(e: Employee) = e.name }
object SalaryToken extends Token[Int] { def value(e: Employee) = e.salary }
object IsDirectorToken extends Token[Boolean] { def value(e: Employee) = e.isDirector }

<强>用法

scala> val john = new Employee("John", 50000, false)
john: Employee = Employee@59f99ea

scala> val name = john.get(NameToken)
name: String = John

scala> val salary = john.get(SalaryToken)
salary: Int = 50000

scala> val isDirector = john.get(IsDirectorToken)
isDirector: Boolean = false

答案 1 :(得分:1)

您可以使用重载:

  case class Employee(name: String, salary: Int, isDirector: Boolean) {
  def get(t: Token[String]) = name;
  def get(t: Token[Int]) = salary;
  def get(t: Token[Boolean]) = isDirector;
  }