3

假设我有两个元组，第一个是类型为值的元组(V1, V2, .., Vn)，

第二个是类型为函数的元组(V1 => V1, V2 => V2, .., Vn => Vn)。

现在我想将这两个元组(f1(v1), v2(v2), .., fn(vn))与 type结合起来(V1, V2, .., Vn)。

scala> val values = (1, 2.0, "3")
val values: (Int, Double, String) = (1,2.0,3)

scala> val funs = ((i: Int) => 2 * i, (f: Double) => 2 * f, (s: String) => s * 2)
val funs: (Int => Int, Double => Double, String => String) = ..

scala> val res = ???  // (2, 4.0, "33")

我不知道如何在scala 3.0（即dotty）中得到这个。

编辑：我查看了 shapeless 的源代码并得到了一个（部分工作）解决方案：

scala> trait Zip[V <: Tuple, F <: Tuple]{ type R <: Tuple; def apply(v: V, f: F): R }

scala> given Zip[Unit, Unit]{ type R = Unit; def apply(v: Unit, f: Unit): Unit = () }

scala> given [Hv, Hr, V <: Tuple, F <: Tuple](given z: Zip[V, F]): Zip[Hv *: V, (Hv => Hr) *: F] = new Zip {
     |   type R = Hr *: z.R
     |   def apply(v: Hv *: V, f: (Hv => Hr) *: F): R = {
     |     f.head(v.head) *: z.apply(v.tail, f.tail)
     |   }
     | }

scala> val values = (1, 2.0, "3")
val values: (Int, Double, String) = (1,2.0,3)

scala> val funs = ((i: Int) => 2 * i, (f: Double) => 2 * f, (s: String) => s * 2)
val funs: (Int => Int, Double => Double, String => String) = ..

scala> def apply[V <: Tuple, F <: Tuple](v: V, f: F)(given z: Zip[V, F]): z.R = z.apply(v, f)
def apply[V <: Tuple, F <: Tuple](v: V, f: F)(given z: Zip[V, F]): z.R

scala> apply(values, funs)
val res0:
  Zip[(Int, Double, String), (Int => Int, Double => Double, String => String)]#R = (2,4.0,33)

scala> val res: (Int, Double, String) = apply(values, funs)
1 |val res: (Int, Double, String) = apply(values, funs)
  |                                 ^^^^^^^^^^^^^^^^^^^
  |Found:    ?1.R
  |Required: (Int, Double, String)
  |
  |where:    ?1 is an unknown value of type Zip[(Int, Double, String), (Int => Int, Double => Double, String => String)]

我不知道为什么apply方法的返回会丢失它的类型。

4

3 回答 3

4

为什么 apply 方法的返回丢失了它的类型

这是因为您丢失了类型细化（这种行为在 Scala 2 和 Dotty 中类似）。

编码

given [Hv, Hr, V <: Tuple, F <: Tuple](given z: Zip[V, F]): Zip[Hv *: V, (Hv => Hr) *: F] = new Zip {
  ...

应该

given [Hv, Hr, V <: Tuple, F <: Tuple](given z: Zip[V, F]): (Zip[Hv *: V, (Hv => Hr) *: F] { type R = Hr *: z.R }) = new Zip[Hv *: V, (Hv => Hr) *: F] {
  ...

或带Aux图案

given [Hv, Hr, V <: Tuple, F <: Tuple](given z: Zip[V, F]): Zip.Aux[Hv *: V, (Hv => Hr) *: F, Hr *: z.R] = new Zip[Hv *: V, (Hv => Hr) *: F] {
  ...

在 0.21.0-RC1 中测试。

于 2020-01-04T16:15:49.357 回答
2

这似乎可以解决问题

val res = List(Range(0, values.productArity).map(n => {
  val arg = values.productElement(n)
  val f = funs.productElement(n).asInstanceOf[(arg.type) => arg.type]
  f.apply(arg)
})).map {
  case Vector(a, b, c) => Tuple3(a, b, c)
}.head
于 2020-01-04T11:18:50.017 回答
2

zipApplyfrom shapeless 将保持类型安全，例如

import shapeless.syntax.std.tuple._

val values = (1, 2.0, "3")
val funs = ((i: Int) => 2 * i, (f: Double) => 2 * f, (s: String) => s * 2)
funs zipApply values

输出

res0: (Int, Double, String) = (2,4.0,33)

但是尝试使用val values = ("1", "2.0", "3")会产生编译时错误。

于 2020-01-04T11:55:47.860 回答