Question

如何为抽象类中的方法提供参数，以便以后可以扩展参数？我将在下面举例说明一个简单的场景。

abstract class Car {
  def drive(x: Driver)
}

abstract class Driver
case class DriverA(name: String) extends Driver
case class DriverB(name: String, age: Int) extends Driver

class Audi extends Car {
  // each child class should have a more specific type for param "x"
  def drive(x: DriverA) = { ... }
}

class BMW extends Car {
  // each child class should have a more specific type for param "x"
  def drive(x: DriverB) = { ... }
}

但这在Scala中不起作用：“错误：[..]方法参数类型必须完全匹配”

我也试过指定上限类型，就像这个def驱动[T＆lt;：Driver]（x：T）：Int，但仍然没有运气。似乎我错过了一些简单而明显的东西。

这样的设计有问题吗？

Answer 1

您可以在scala中使用abstract type：

scala> :paste
// Entering paste mode (ctrl-D to finish)

abstract class Car {
  type D <: Driver
  def drive(x: D): Unit
}
abstract class Driver
case class DriverA(name: String) extends Driver
case class DriverB(name: String) extends Driver
class Audi extends Car {
  type D = DriverA
  def drive(x: DriverA): Unit = println(x.name)
}

// Exiting paste mode, now interpreting.

defined class Car
defined class Driver
defined class DriverA
defined class DriverB
defined class Audi

scala> new Audi().drive(DriverA("test"))
test

或者使用类型为参数化的类：

scala> :paste
// Entering paste mode (ctrl-D to finish)

abstract class Driver
case class DriverA(name: String) extends Driver
abstract class Car[D <: Driver] {
  def drive(x: D): Unit
}
class Audi extends Car[DriverA] {
  def drive(x: DriverA) = println(x.name)
}

// Exiting paste mode, now interpreting.

scala> new Audi().drive(DriverA("zzz"))
zzz

Scala - 可以扩展的抽象基类中的参数类型

1 个答案: