如何在scalaz中自定义monad变压器

Question

这些天我正在学习scalaz。当将scalaz与Haskell进行比较时，我发现很难自定义自己的monad转换器。

我可以在Haskell中执行的代码如下：

newtype Box a = Box a deriving Show
-- Monad instance of Box, it's simple, I pass it..
newtype BoxT m a = BoxT {runBoxT :: m (Box a)}

instance (Monad m) => Monad (BoxT m) where
    return a = BoxT $ return $ Box a
    x >>= f = BoxT $ do
        (Box v) <- runBoxT x
        runBoxT $ f v

instance (Monad m) => Applicative (BoxT m) where
    pure = return
    (<*>) = ap

instance (Monad m) => Functor (BoxT m) where
    fmap = liftM

我如何在scalaz中做同样的事情？我已经尝试了很多方法，但是似乎无法弄清楚scala的泛型类型，如下所示：

case class Box[A](value:A)
case class BoxT[M[_],A](run: M[Box[A]])

implicit val monad = Monad[BoxT] { // goes wrong, type `BoxT[M[_],A]` doesn't match F[_]
 ... //don't know how to override functions..
}

Answer 1

感谢@Iva Kam对同类投影仪的评论。

首先，在build.sbt中启用了种类的投影仪：https://github.com/typelevel/kind-projector

addCompilerPlugin("org.typelevel" %% "kind-projector" % "0.11.0" cross CrossVersion.full)

然后代码可以像这样工作：

  case class Box[A](value:A)
  case class BoxT[M[_],A](run: M[Box[A]])

  implicit def monad[M[_]](implicit m : Monad[M]): Monad[BoxT[M, ?]] = new Monad[BoxT[M,?]] {
    override def point[A](a: => A): BoxT[M,A] = BoxT(m.point(Box(a)))

    override def bind[A, B](fa: BoxT[M,A])(f: A => BoxT[M,B]): BoxT[M,B] = BoxT(m.bind(fa.run){ b: Box[A] =>
      val v = b.value
      f(v).run
    })
  }

  def main(args: Array[String]): Unit = {
    val a = Monad[BoxT[Option,?]].point(2)
    val b = for {
      v <- a  //v should be a Int, but idea can't analyze it, maybe I should update, still in 2018.2
    } yield {v * 4}

    val c = a >> b //this line works, but idea can't analyze it :(
    println(b)
    println(c)
  }

但是看起来当我使用BoxT的Monad时，我应该使用?而不是Box内部类型的A。