Scala 3 中的元组类型类派生的编译器操作

Question

学习一些新的 Scala 3 编译时操作并对

Tuple

有点困惑（特别是在

*:

和

EmptyTuple

上使用类型匹配）

import scala.compiletime.*
imort cats.Show

transparent inline def showForTuple[T <: Tuple]: Show[T] =
  inline erasedValue[T] match
    case _: EmptyTuple => (new Show[EmptyTuple] {
      override def show(n: EmptyTuple): String = ""
    }).asInstanceOf[Show[T]]

    case _: (t *: EmptyTuple) => (new Show[t *: EmptyTuple] {
      val showHead = summonInline[Show[t]]

      override def show(tup: t *: EmptyTuple): String = showHead.show(tup.head)
    }).asInstanceOf[Show[T]]

    case _: (t *: ts) => (new Show[t *: ts] {
      val showHead = summonInline[Show[t]]
      val showTail = showForTuple[ts]

      override def show(tup: t *: ts): String =
        showHead.show(tup.head) + ", " + showTail.show(tup.tail)
    }).asInstanceOf[Show[T]]

这在 Scala 3.2.2 上按预期工作：

showForTuple[Int *: String *: EmptyTuple].show((1, "hola mundo"))
val res2: String = 1, hola mundo

但是失败了：

showForTuple[(Int, String)].show((1, "hola mundo"))
java.lang.VerifyError: Bad type on operand stack
Exception Details:
  Location:
    rs$line$28$$anon$1.show(Lscala/Product;)Ljava/lang/String; @16: invokevirtual
  Reason:
    Type 'scala/Product' (current frame, stack[2]) is not assignable to 'scala/Tuple2'
  Current Frame:
    bci: @16
    flags: { }
    locals: { 'rs$line$28$$anon$1', 'scala/Product', 'scala/Product' }
    stack: { 'java/lang/StringBuilder', 'cats/Show', 'scala/Product' }
  Bytecode:
    0000000: bb00 2c59 122d b700 302a b600 322b 4d2c
    0000010: b600 38b8 003e b800 42b9 0045 0200 b600
    0000020: 4912 4bb6 0049 2ab6 004d 2b4e b200 522d
    0000030: b600 55b6 0059 b900 4502 00b6 0049 b600
    0000040: 5cb0                                   

  ... 66 elided

学习自：https://docs.scala-lang.org/scala3/reference/metaprogramming/compiletime-ops.html 和其他资源（博客文章/视频）

---

编辑：感谢 Scala 的 Discord 中的 Il Totore，以下是一种解决方法，但对最初尝试时导致 java.lang.VerifyError 的原因感到困惑：

import scala.compiletime.*
import cats.Show

given Show[EmptyTuple] = _ => ""
lazy val given_Show_EmptyTuple: cats.Show[EmptyTuple]

given [A, T <: Tuple](using showA: Show[A], showT: Show[T]): Show[A *: T] =
  _ match
    case h *: EmptyTuple => showA.show(h)
    case h *: t => showA.show(h) + ", " + showT.show(t)

transparent inline def showForTuple[T <: Tuple]: Show[T] =
  inline erasedValue[T] match
    case _: EmptyTuple => summonInline[Show[EmptyTuple]].asInstanceOf[Show[T]]
    case _: (t *: EmptyTuple) => summonInline[Show[t *: EmptyTuple]].asInstanceOf[Show[T]]
    case _: (t *: ts) => summonInline[Show[t *: ts]].asInstanceOf[Show[T]]

showForTuple[Int *: String *: EmptyTuple].show((1, "hola mundo")) // works
showForTuple[(Int, String)].show((1, "hola mundo")) // Also works?

Answer 1

这是使用内置方法

scala.compiletime.summonAll

和类型级操作

Tuple.Map

、

Zip

等的另一种解决方法

type Ev[T <: Tuple, A] = Tuple.Union[Tuple.Map[T, [_] =>> A]] =:= A

inline def showForTuple[T <: Tuple]: Show[T] =
  new Show[T]:
    override def show(t: T): String =
      summonFrom {
        case ev: Ev[Tuple.Zip[Tuple.Map[T, Show], T], String] =>
          ev.liftCo[List](
            summonAll[Tuple.Map[T, Show]]
              .zip(t)
              .map[[_] =>> String]([a] => (x: a) =>
                type InstVal[b] = (Show[b], b)
                x match
                  case (s, v): InstVal[?] => s.show(v)
              ).toList
          ).reduce(_ + ", " + _)
      }

我不得不使用

scala.compiletime.summonFrom

因为编译器不知道

Tuple.Union[Tuple.Map[T, [_] =>> A]] =:= A

.

更简单的实现是用

mkString

代替

reduce

inline def showForTuple[T <: Tuple]: Show[T] =
  new Show[T]:
    override def show(t: T): String =
      summonAll[Tuple.Map[T, Show]]
        .zip(t)
        .map[[_] =>> String]([a] => (x: a) =>
          type InstVal[b] = (Show[b], b)
          x match
            case (s, v): InstVal[?] => s.show(v)
        )
        .toList
        .mkString(", ")