如何使用 shapeless 连接函数参数和返回值

问题描述

我试图在下面定义 compose 方法：

import shapeless.{::,HList,HNil}

def compose[A1 <: HList,R1,A2 <: HList,R2](f: A1 => R1,g: A2 => R2): R1 :: R2 :: HNil = ???

val f = (xs: Int :: String :: HNil) => xs.select[Int].toString + xs.select[String]
val g = (xs: Boolean :: HNil) => xs.select[Boolean]

// expected output:
// Int :: String :: Boolean :: HNil => String :: Boolean :: HNil
compose(f,g)

这是我的不完整代码，我不知道如何从 A1 获取 A2 和 args。有谁可以帮忙吗？

def compose[A1 <: HList,g: A2 => R2)(implicit p: Prepend[A2,A1]) =
  (args: p.Out) => {
    val a1: A1 = ???
    val a2: A2 = ???
    f(a1) :: f(a2) :: HNil
  }

解决方法

我能够使用 Split 实施一个可行的解决方案：

def compose[A1 <: HList,R1,A2 <: HList,R2,O <: HList,N <: Nat](f: A1 => R1,g: A2 => R2)(
  implicit split: Split.Aux[O,N,A1,A2]
): O => R1 :: R2 :: HNil = {
  (a: O) =>
    val (a1,a2) = split.apply(a)
    f(a1) :: g(a2) :: HNil //I assumed you meant calling g(a2) here
}

val f = (xs: Int :: String :: HNil) => xs.select[Int].toString + xs.select[String]
val g = (xs: Boolean :: HNil) => xs.select[Boolean]

val r: Int :: String :: Boolean :: HNil => String :: Boolean :: HNil = compose(f,g)

println(r.apply(1 :: "hello" :: true :: HNil)) // 1hello :: true :: HNil