scala - 如何替换通用匿名函数？

Question

假设有腿动物有一个特征:

trait Legged {
  val legs: Int

  def updateLegs(legs: Int): Legged
}

有两种这样的有腿的动物:

case class Chicken(feathers: Int, legs: Int = 2) extends Legged {
  override def updateLegs(legs: Int): Legged = copy(legs = legs)
}

case class Dog(name: String, legs: Int = 4) extends Legged {
  override def updateLegs(legs: Int): Legged = copy(legs = legs)
}

在农场里还有这些动物的持有人

case class Farm(chicken: Chicken, dog: Dog)

以及一种通过添加一条额外的腿来变异所有有腿动物的通用方法

def mutate(legged: Legged): Legged = legged.updateLegs(legged.legs + 1)

问题是如何在Farm上实现一个方法所以它需要 mutate: Legged => Legged函数作为参数并将其应用于所有 Legged动物？

val farm = Farm(Chicken(1500), Dog("Max"))
farm.mapAll(mutate) //this should return a farm whose animals have an extra leg

到目前为止我所带来的，但实际上并没有用

trait LeggedFunc[T <: Legged] extends (T => T)


case class Farm(chicken: Chicken, dog: Dog) {
  def mapAll(leggedFunc: LeggedFunc[Legged]): Farm = {
    //todo how to implement?
    val c = leggedFunc[Chicken](chicken)
  }
}

我知道如何使用模式匹配来做到这一点，但这会导致潜在的 MatchError .

Answer 1

一种可能的方法(类型安全，不使用 asInstanceOf )可能是使用依赖于对象的类型。
首先，我们应该添加一个抽象成员，它使用 Legged 的具体类型。子类:

sealed trait Legged { self =>
  type Me >: self.type <: Legged // F-Bounded like type, Me have to be the same type of the subclasses
  val legs: Int
  def updateLegs(legs: Int): Me
}

然后，Legged子类变成了:

case class Chicken(feathers: Int, legs: Int = 2) extends Legged {
  type Me = Chicken
  override def updateLegs(legs: Int): Chicken = copy(legs = legs)
}

case class Dog(name: String, legs: Int = 4) extends Legged {
  type Me = Dog
  override def updateLegs(legs: Int): Dog = copy(legs = legs)
}

这样，就可以定义一个返回Legged的具体子类的函数。通过(类似于@Gaël J 所做的):

trait LeggedFunc {
  def apply(a : Legged): a.Me
}

val mutate = new LeggedFunc { override def apply(legged: Legged): legged.Me = legged.updateLegs(legged.legs + 1) }

最后，Farm类很简单，定义为:

case class Farm(chicken: Chicken, dog: Dog) {
  def mapAll(leggedFunc: LeggedFunc): Farm = {
    val c : Chicken = leggedFunc(chicken)
    val d : Dog = leggedFunc(dog)
    Farm(c, d)
  }
}

Scastie对于 Scala 2
但是为什么是对象依赖类型呢？
在 Scala 3.0 中，可以定义 dependent function type作为:

type LeggedFunc = (l: Legged) => l.Me
val mutate: LeggedFunc = (l) => l.updateLegs(l.legs + 1)

使这个解决方案(依赖于对象的类型)更清晰和类型安全。
Scastie对于 Scala 3 版本