Clarification on enums with fields

[olafurpg]

Related to #5221

One question about enums, https://dotty.epfl.ch/docs/reference/enums/adts.html

enum Option[+T] {
  case Some(x: T)
  case None
}

Is the Option.Some type visible? Is it legal to reference it like this def get(some: Option.Some[Int]) = some.x?

[OlivierBlanvillain]

As it is currently implemented, yes, this type is visible. It's also possible to explicitly call the constructor to get instances:

scala> enum Option[+T] {
     |   case Some(x: T)
     |   case None
     | }
// defined class Option

scala> new Option.Some(1)
val res0: Option.Some[Int] = Some(1)

But you only get a class (and a type) for cases with fields. Cases without fields become vals:

scala> new Option.None
1 |new Option.None
  |    ^^^^^^^^^^^
  |    type `None` is not a member of Option - did you mean `Option.None`?

scala> type nope = Option.None
1 |type nope = Option.None
  |            ^^^^^^^^^^^
  |       type `None` is not a member of Option - did you mean `Option.None`?

scala> type nope = Option.None.type
// defined alias type nope = Option[Nothing](Option.None)

[olafurpg]

Thanks! I just realized this is already addressed in http://dotty.epfl.ch/docs/reference/enums/adts.html

Note that the type of the expressions above is always Option. That is, the implementation case classes are not visible in the result types of their apply methods. This is a subtle difference with respect to normal case classes. The classes making up the cases do exist, and can be unveiled by constructing them directly with a new.

I apologize the noise.