Bifunctors 的参数化类型别名

Question

我有一个 Seq[R]，我想将其拆分为一个 Tuple2[Seq[E], Seq[S]]，在编写代码时，我想到了一个事实，即我可以为元组使用自定义 Bifunctor seqs 和作为练习尝试编写代码：

import scalaz.Bifunctor

type MyType[E, S] = (Seq[E], Seq[S])

case class MyVali[E, S](tp: (Seq[E], Seq[S]))(implicit bifunctor: Bifunctor[MyType]) {
  def bimap[C, D](f: (E) => C, g: (S) => D): (Seq[C], Seq[D]) =
    bifunctor.bimap(tp)(f, g)

  def leftMap[C](f: (E) => C): (Seq[C], Seq[S]) =
    bifunctor.leftMap(tp)(f)

  def rightMap[D](g: (S) => D): (Seq[E], Seq[D]) =
    bifunctor.rightMap(tp)(g)

}

val myValBifunctorInstance = new Bifunctor[MyType] {
  override def bimap[A, B, C, D](fab: (Seq[A], Seq[B]))(f: (A) => C, g: (B) => D): (Seq[C], Seq[D]) =
    (fab._1.map(f), fab._2.map(g))
}

MyVali((Seq.empty[String], Seq.empty[Int]))(myValBifunctorInstance).bimap(a => a, b => b)

这很好用，但出于某种原因我不明白，我不得不声明一个参数化类型别名来编译所有这些，即 type MyType[E, S] = (Seq[E], Seq[S]) 我几乎不明白为什么这有效而这不起作用：

def myValBimap[E, S] = new Bifunctor[Tuple2[Seq[E], Seq[S]]] {
  override def bimap[A, B, C, D](fab: (A, B))(f: (A) => C, g: (B) => D): (C, D) = ???
}

[error] ... (Seq[E], Seq[S]) takes no type parameters, expected: two

[error] def myValBimap[E, S] = new Bifunctor[Tuple2[Seq[E], Seq[S]]] {

当定义了这样的类型别名时，编译器是否会创建 2 类型的挂起（可能像嵌套的 lambda 类型？）？

Answer 1

def myValBimap[E, S] = new Bifunctor[Tuple2[Seq[E], Seq[S]]] {  ... }

上面Bifunctor中的Tuple2[...]没有两个类型参数了，因为E和S被填入了

例如 myValBimap[Int, String] 试图创建一个 Bifunctor[(Seq[Int], Seq[String])] 并且类型 (Seq[Int], Seq[String]) 显然没有两个类型参数。

你可以写

Bifunctor[({ type λ[α, β] = (Seq[α], Seq[β])})#λ] 或
Bifunctor[λ[(α, β) => (Seq[α], Seq[β])]]使用kind projector plugin.

如果你需要一个额外的类型，你需要一个带有类型参数的隐式函数，可能是这样的：

implicit def myBimap[F[_]: Functor]: Bifunctor[λ[(α, β) => (F[α], F[β])]] = ???

一个更简单的例子是 Functor for disjunction / \/ 它使用左侧的类型参数：

implicit def functorDisj[L]: Functor[L \/ ?] = ???
implicit def functorDisj[L]: Functor[({ type λ[α] = L \/ α })#λ] = ???

关于您的原始问题：可以将您的 R 映射到 E \/ S 并使用 separate :

import scalaz._, Scalaz._

val integers = List.range(1,10)
val stringOrInts: List[String \/ Int] = 
  integers.map(i => if (i % 2 == 0) i.right else i.toString.left)

stringOrInts.separate
// (List[String], List[Int]) = (List(1, 3, 5, 7, 9),List(2, 4, 6, 8))

Bifunctors 的参数化类型别名

Parametrized type alias for Bifunctors

scala

functor

scalaz

type-alias

bifunctor