Scala - 可以扩展的抽象基类中的参数类型

Question

如何为抽象类中的方法提供参数，以便稍后可以扩展该参数？我将在下面说明一个简单的情况。

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) = { ... } 
} 

但是，这并不Scala中工作：“错误：[..]方法参数类型必须完全匹配”

我还试图指定上限类型这样DEF驱动[T <：驱动]（x：T）：Int，但仍然没有任何运气。看来我缺少一些简单而明显的东西。

这样的设计有什么问题吗？

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 

或者使用类参数类型：

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