在scala中获取单例类型的实例

Question

我想在Scala中获得单例类型的实例，这可能吗？

示例：（我知道在这种情况下可以更轻松地完成）

sealed trait Animal
case object Dog extends Animal
case object Cat extends Animal

trait Person {
  def name: String
  // Is there a "FavoriteAnimal.instance" syntax?
  def mostImportantThings = (FavoriteAnimal.instance, name)
  protected type FavoriteAnimal <: Animal with scala.Singleton
}

case class DogPerson(override val name: String) extends Person {
  override type FavoriteAnimal = Dog.type
}

case class CatPerson(override val name: String) extends Person {
  override type FavoriteAnimal = Cat.type
}

Answer 1

也许你想要类似的东西

sealed trait Animal
case object Dog extends Animal
case object Cat extends Animal

trait Person[A <: Animal] {
  def name: String
  def animal: A
  def mostImportantThings = (animal, name)
}

case class DogPerson(override val name: String) extends Person[Dog.type] {
  override val animal = Dog
}

case class CatPerson(override val name: String) extends Person[Cat.type] {
  override val animal = Cat
}

Answer 2

使用shapeless.Witness正确的语法是

sealed trait Animal
case object Dog extends Animal
case object Cat extends Animal

trait Person {
  def name: String
  def mostImportantThings(implicit 
    witness: Witness.Aux[FavoriteAnimal]
  ): (FavoriteAnimal, String) = (witness.value, name)
  protected type FavoriteAnimal <: Animal with scala.Singleton
}

case class DogPerson(override val name: String) extends Person {
  override type FavoriteAnimal = Dog.type
}

case class CatPerson(override val name: String) extends Person {
  override type FavoriteAnimal = Cat.type
}

DogPerson("A Dog Person").mostImportantThings // (Dog, A Dog Person)

不幸的是，在当前版本的Shapeless（2.3.3）中，存在一个错误，并且该代码无法编译。 但是修复之后就可以了。