scala – 将构造函数作为返回类型

scala> sealed trait Maybe[A]
defined trait Maybe

scala> case class Just[A](x: A) extends Maybe[A]
defined class Just

scala> case object NothingHere extends Maybe[Nothing]
defined object NothingHere

可以使用返回类型Maybe [A]返回函数f.

scala> def f[A](x: A): Maybe[A] = Just(x)
f: [A](x: A)Maybe[A]

但是,也可以指定返回Just [A].

scala> def f[A](x: A): Just[A] = Just(x)
f: [A](x: A)Just[A]

现在我将在Haskell中进行类似的练习：

Prelude> data Option a = None | Some a deriving Show
Prelude> let f x = Some x :: Option Int
Prelude> f 10
Some 10

但是,我无法设置类型构造函数的返回类型.

Prelude> let f x = Some x :: Some Int

<interactive>:10:21:
    Not in scope: type constructor or class `Some'
    A data constructor of that name is in scope; did you mean DataKinds?
Prelude> let f x = None :: None

Scala的Just是一个类的简单区别,即合法的返回类型吗？然而,在Haskell中,类型构造函数不能是返回类型？