How can I alias a covariant generic type parameter

Question

The following code does not compile (in Scala 2.11):

case class CovariantClass[+R](value: R) {
  type T = R
  def get: R = value
}

object Main {
  def main(args: Array[String]): Unit ={
    println(CovariantClass[String]("hello").get)
  }
}

The error message is:

Error:(4, 8) covariant type R occurs in invariant position in type R of type T
  type T = R
       ^

Why can't I alias a covariant type parameter? If I remove the line type T = R , the code compiles and prints hello , so the alias seems to be the problem. Unfortunately, this means that I cannot create an alias for more complex types, eg, type T = List[R] does not compile either, although List is covariant.

Answer 1

From the scala spec :

The right-hand side of a type alias is always in invariant position.

This means that you cannot create the alias T and specify a variant type R on the right-hand side. The same applies to List[R] , because it is also covariant.

You can , however provide a type alias with a type parameter:

case class CovariantClass[+R](value: R) {
  type T[+R] = List[R]
  def get: R = value
}

If you find yourself wanting to alias the type parameter R , you should probably just name it something else in the first place.

Answer 2

It's forbidden because it would allow a program that isn't correct, that's always the rule. You could rewrite it like this:

case class CovariantClass[+R](value: R) {
  type T <: R
  def get: R = value
}

As for an example of how it breaks, consider this:

case class CovariantClass[+R](value: R) {
  type T = Int
  def get: R = value
  def put(x: T) {}
  def put2(x: R) {}
}

Because of how T is defined, it is invariant. That means it can be used in places where covariant types cannot, such as seen above. Notice that put compiles but put2 does not.