如何组合具有兼容类型的两个元组？

Question

假设我有两个元组，第一个是类型为 (V1, V2, .., Vn)、

的值元组

第二个是类型为 (V1 => V1, V2 => V2, .., Vn => Vn).

的函数元组

现在我想将两个元组合并为 (f1(v1), v2(v2), .., fn(vn)) 类型 (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 中获得它（即 dotty）。

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 return 丢失了它的类型。

Answer 1

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

Answer 2

来自 shapeless 的

zipApply 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

产出

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。

Answer 3

why the return of apply method lost its type

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

代码

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

应该是

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] {
  ...

或使用 Aux 模式

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。

如何组合具有兼容类型的两个元组？

How to combine two tuples with compatible types?

scala

tuples

dotty

shapeless

scala-3