Scala宏如何將`HList`轉換為函數args

Question

對於以下類型

type HFunc = (Int :: String :: HNil) => Int

type Func = (Int, String) => Int

我嘗試將Func轉換為HFunc

val funExpr: Tree = ???
val hlistType = ???      
val hfuncName = c.freshName("hfunc")

q"""
  def $hfuncName(t: $hlistType) = {
    ${funExpr}(..) //how to extract hlist elements as params ?
  }
"""

如何提取HList元素並將其傳遞給Func ？

Answer 1

如果您只想進行轉換（不一定是宏），那么shapeless可以將它們作為函數的擴展方法提供：

import shapeless._
import shapeless.syntax.std.function._

type HFunc = (Int :: String :: HNil) => Int
type Func = (Int, String) => Int

def toHFunc(f: Func): HFunc = f.toProduct
def fromHFunc(hf: HFunc): Func = hf.fromProduct