“MyType” problem: Do I have to use abstract types (or generics) in Scala to return the actual class?

Question

I am not sure if there is a better way of doing this:

trait Animal {
  val name: String
  val weight: Int

  type SubAnimal <: Animal

  def updateName(n:         


        

Answer 1

This should work:

trait Animal[T] {
  self:T =>

  val name: String
  val weight: Int

  def updateName(n: String): T = returnMe(n, this.weight)
  def updateWeight(w: Int): T = returnMe(this.name, w)
  // Abstract protected method
  protected def returnMe(n: String, w: Int): T
}

case class Dog(name: String, weight: Int) extends Animal[Dog] {
  override def returnMe(n: String, w: Int): Dog = Dog("Dog: " + name, w)
}

case class Cat(name: String, weight: Int) extends Animal[Cat] {
   override def returnMe(n: String, w: Int): Cat = Cat("Cat: " + name, w)
}

Something like case class Cat(name: String, weight: Int) extends Animal[Dog] gets rejected by the compiler. Code stolen adapted from http://oldfashionedsoftware.com/2009/12/10/self-help/

Answer 2

Using type parametrization?

trait Animal[A <: Animal[A]] {
  val name: String
  val weight: Int

  def updateName(n: String) = returnMe(n, this.weight)
  def updateWeight(w: Int) = returnMe(this.name, w)
  // Abstract protected method
  protected def returnMe(n: String, w: Int): A
}

case class Dog(name: String, weight: Int) extends Animal[Dog] {
  override def returnMe(n: String, w: Int) = Dog("Dog: " + name, w)
}

case class Cat(name: String, weight: Int) extends Animal[Cat] {
  override def returnMe(n: String, w: Int) = Cat("Cat: " + name, w)
}

Answer 3

Some recent discussion on this topic... What you're looking for is commonly referred to as "MyType", and the typical Scala/Java encoding for it uses recursive type parameters:

public abstract class Enum<E extends Enum<E>> implements Comparable<E>, Serializable { 
    public final int compareTo(E o) 
    // ... 
}