scala - 多态方法适用于 lambda 类型，但不适用于 Scala 3 中的类型通配符

Question

在 Scala 3 中，我可以使用 lambda 类型为状态定义仿函数:

  given stateFunctor[S]: Functor[[A] =>> State[S, A]] with
    override def map[A, B](a: State[S, A])(fx: A => B): State[S, B] = State(a.run.andThen { case (s, a) => (s, fx(a)) })

我希望它能与 ? 或 _ 通配符一起使用:

  given stateFunctor[S]: Functor[State[S, ?]] with
    override def map[A, B](a: State[S, A])(fx: A => B): State[S, B] = State(a.run.andThen { case (s, a) => (s, fx(a)) })

但我收到以下编译错误:

Type argument domain.State[S, ? <: AnyKind] does not have the same kind as its bound [_$1] given stateFunctor[S]: Functor[State[S, ? <: AnyKind]] with

为什么不起作用？我缺少什么？我认为 Scala 3 支持类型通配符的种类投影语法。

Scala 版本:3.1.3

如果您需要它，这里是State和Functor定义:

case class State[S, A](run:S => (S, A)):
  def exec(s:S):S = run(s)._1
  def eval(s:S):A = run(s)._2

trait Functor[F[_]]:
  def map[A, B](a: F[A])(fx: A => B): F[B]

Answer 1

? 错误。 ? 用于存在类型 State[S, ?] (在 Scala 2 中是 State[S, _] 又名 State[ S, A] forSome { type A })，不适用于 lambda 类型。

_ 代表 lambda 类型(在 Scala 2 中，它们被模拟 ({ type F[A] = State[S, A] })#F)。所以它应该是 State[S, _] 但这尚未实现。

https://docs.scala-lang.org/scala3/reference/changed-features/wildcards.html

The syntax of wildcard arguments in types has changed from _ to ?

We would like to use the underscore syntax _ to stand for an anonymous type parameter, aligning it with its meaning in value parameter lists. So, just as f(_) is a shorthand for the lambda x => f(x), in the future C[_] will be a shorthand for the type lambda [X] =>> C[X].

到目前为止，您可以编写 [A] =>> State[S, A] 或使用 kind projector 状态[S，*]

scalaVersion := "3.2.1"

scalacOptions += "-Ykind-projector"