scala - Scala 中带有两个参数的类型构造函数的仿函数实例

Question

我有一个类Foo有两个参数，我正在尝试为第一个参数固定的 Foo 编写一个 Functor 实例，如下所示:

object Scratchpad {

  trait Functor[F[_]] {
    def fmap[A, B](f: A => B): F[A] => F[B]
  }

  case class Foo[X, Y](value: Y)

  implicit def fooInstances[X]: Functor[Foo[X, _]] =
    new Functor[Foo[X, _]] {
      def fmap[A, B](f: A => B): Foo[X, A] => Foo[X, B] =
        foo => Foo[X, B](f(foo.value))
    }
}

但是上面的代码编译失败，产生如下错误:

Error:(9, 41) Scratchpad.Foo[X, _] takes no type parameters, expected: one
  implicit def fooInstances[X]: Functor[Foo[X, _]] =

我知道 Scalaz 用他们的 \/ 做了类似的事情类型，但检查他们的源代码发现一个奇怪的 ? ，它不会为我编译:

implicit def DisjunctionInstances1[L]: Traverse[L \/ ?] with Monad[L \/ ?] with BindRec[L \/ ?] with Cozip[L \/ ?] with Plus[L \/ ?] with Optional[L \/ ?] with MonadError[L \/ ?, L] =

Scalaz 怎么样?工作，以及如何为 Foo 编写 Functor 实例?

Answer 1

but inspection of their source code reveals an odd ?, which doesn't compile for me

?来自 kind-projector 项目，这是一个 Scala 编译器插件，您需要添加到您的 build.sbt :

resolvers += Resolver.sonatypeRepo("releases")

addCompilerPlugin("org.spire-math" %% "kind-projector" % "0.9.4")

这为您美化了类型 lambda 的创建:

implicit def fooInstances[X]: Functor[Foo[X, ?]] =
  new Functor[Foo[X, ?]] {
    def fmap[A, B](f: A => B): Foo[X, A] => Foo[X, B] =
      foo => Foo[X, B](f(foo.value))
}

请记住，我们还可以使用具有类型别名的部分类型应用程序:

implicit def fooInstances[X] = {
  type PartiallyAppliedFoo[A] = Foo[X, A]
  new Functor[PartiallyAppliedFoo] {
    override def fmap[A, B](f: (A) => B): (PartiallyAppliedFoo[A]) => PartiallyAppliedFoo[B] = foo => Foo[X, B](f(foo.value))
  }
}

编辑 (05/04/2020)

注意 kind 项目的语法从 ? 改变了。至 *对于部分应用类型，因此上面的示例应该是:

sbt:

resolvers += Resolver.sonatypeRepo("releases")

addCompilerPlugin("org.spire-math" %% "kind-projector" % "0.11.0")

代码:

implicit def fooInstances[X]: Functor[Foo[X, *]] =
  new Functor[Foo[X, *]] {
    def fmap[A, B](f: A => B): Foo[X, A] => Foo[X, B] =
      foo => Foo[X, B](f(foo.value))
}