如果我要隐式地将两个对象从一个转换为另一个,是否可以使用Iso宏之类的方法来执行此操作?
例如,如果我有这个:
implicit def listToMap[A, B](l: List[(A, B)]): Map[A, B] = l.toMap
implicit def mapToList[A, B](m: Map[A, B]): List[(A, B)] = m.toList
我想简单地写:
implicit def[A, B] listMapIso = Iso[List[(A, B)], Map[A, B]] {_.toMap, _.toList}
注意:如下所述,我计划在Web框架中使用此模型,在该框架中将数据库模型转换为中间件/前端模型。
最佳答案
您似乎在混淆几个不同的概念。同样,隐式转换和宏都彼此完全不同。
我们当然可以为参数化类型定义Iso的等效项,尽管语法变得有些麻烦:
import scalaz._, Scalaz._
case class BiIso[F[_, _], G[_, _]](left: F ~~> G,
right: G ~~> F)
type PairList[A, B] = List[(A, B)]
val listToMap = new (PairList ~~> Map) {
def apply[A, B](l: PairList[A, B]) = l.toMap
}
val mapToList = new (Map ~~> PairList) {
def apply[A, B](m: Map[A, B]) = m.toList
}
val listMapIso = BiIso(listToMap, mapToList)
尽管这是一个正交的问题,我们当然可以将其部分隐式化。我们可以隐式构建BiIso:
implicit val listToMap = new (PairList ~~> Map) {
def apply[A, B](l: PairList[A, B]) = l.toMap
}
implicit val mapToList = new (Map ~~> PairList) {
def apply[A, B](m: Map[A, B]) = m.toList
}
implicit def biIso[F[_, _], G[_, _]](implicit left: F ~~> G, right: G ~~> F) =
BiIso(left, right)
implicitly[BiIso[PairList, Map]]
我们可以使任何BiIso都充当隐式转换,尽管我建议不要这样做。唯一棘手的部分是正确地引导类型推断。这是大多数方法,但是由于某些原因,无法推断出GAB参数(非常欢迎进行更正):
sealed trait BiAny[F[_, _]] {}
object BiAny {
implicit def any[F[_, _]] = new BiAny[F] {}
}
sealed trait ApplyBiIso[FAB, GAB] {
type A1
type B1
type F[_, _]
type G[_, _]
type Required = BiIso[F, G]
val unapplyL: Unapply2[BiAny, FAB] {
type A = A1; type B = B1;
type M[C, D] = F[C, D]
}
val unapplyR: Unapply2[BiAny, GAB] {
type A = A1; type B = B1;
type M[C, D] = G[C, D]
}
def liftBI(bi: Required): Iso[FAB, GAB] =
Iso({ fab: FAB =>
val f: F[A1, B1] = Leibniz.witness(unapplyL.leibniz)(fab)
val g: G[A1, B1] = bi.left(f)
Leibniz.witness(Leibniz.symm[⊥, ⊤, GAB, G[A1, B1]](unapplyR.leibniz))(g): GAB
},
{ gab: GAB =>
val g: G[A1, B1] = Leibniz.witness(unapplyR.leibniz)(gab)
val f: F[A1, B1] = bi.right(g)
Leibniz.witness(Leibniz.symm[⊥, ⊤, FAB, F[A1, B1]](unapplyL.leibniz))(f): FAB
}
)
}
object ApplyBiIso {
implicit def forFG[FAB, A2, B2, GAB, A3, B3](
implicit u1: Unapply2[BiAny, FAB] { type A = A2; type B = B2 },
u2: Unapply2[BiAny, GAB] { type A = A3; type B = B3 }) = new ApplyBiIso[FAB, GAB] {
type A1 = A2
type B1 = B2
type F[C, D] = u1.M[C, D]
type G[C, D] = u2.M[C, D]
//Should do the conversion properly with Leibniz but I can't be bothered
val unapplyL = u1.asInstanceOf[Unapply2[BiAny, FAB] {
type A = A1; type B = B1;
type M[C, D] = F[C, D]
}]
val unapplyR = u2.asInstanceOf[Unapply2[BiAny, GAB] {
type A = A1; type B = B1;
type M[C, D] = G[C, D]
}]
}
type Aux[FAB, GAB, Required1] = ApplyBiIso[FAB, GAB] { type Required = Required1 }
def apply[FAB, GAB](implicit abi: ApplyBiIso[FAB, GAB]): Aux[FAB, GAB, abi.Required] = abi
}
sealed trait AppliedBiIso[FAB, GAB] {
val iso: Iso[FAB, GAB]
}
object AppliedBiIso {
implicit def applyAndIso[FAB, GAB, Required1](
implicit ap: ApplyBiIso.Aux[FAB, GAB, Required1],
iso1: Required1) = new AppliedBiIso[FAB, GAB] {
//Should do the conversion properly with Leibniz but I can't be bothered
val iso = ap.liftBI(iso1.asInstanceOf[BiIso[ap.F, ap.G]])
}
}
implicit def biIsoConvert[FAB, GAB](
f: FAB)(implicit ap: AppliedBiIso[FAB, GAB]): GAB =
ap.iso.left(f)
val map: Map[String, Int] = Map("Hello" -> 4)
val list: PairList[String, Int] =
biIsoConvert[Map[String, Int], PairList[String, Int]](map)
毫无疑问,有可能使这项工作正常进行。
仍然留下宏,宏又或多或少是正交的。我可以看到它们可能相关的一个地方是,如果不使用宏,就不可能在scala中抽象出种类。您是否想要一个等效的Iso,它适用于任何“形状”,而不仅仅是
F[_, _]
?那将是一个宏的好用例-尽管在我不羡慕任何尝试实现它的人之前编写了这种宏。