使用多态函数映射泛型元组

Scala 3 提供了类似于 shapeless 的多态函数和元组HList：

scala> 1 *: "foo" *: Tuple()
val res0: (Int, String) = (1,foo)

scala> val f: ([T] => T => Option[T]) = [T] => (v: T) => Some(v)
val f: PolyFunction{apply: [T](x$1: T): Option[T]} = <function1>

scala> res0.map(f)
val res1: Option[Int] *: Option[String] *: EmptyTuple = (Some(1),Some(foo))

我们如何使用 Scala 3 功能重新实现以下无形示例？

import poly._

object choose extends (Set ~> Option) {
  def apply[T](s : Set[T]) = s.headOption
}

scala> val sets = Set(1) :: Set("foo") :: HNil
sets: Set[Int] :: Set[String] :: HNil = Set(1) :: Set(foo) :: HNil

scala> val opts = sets map choose
opts: Option[Int] :: Option[String] :: HNil = Some(1) :: Some(foo) :: HNil

换句话说，我们怎么能让这样的东西编译

scala> val choose: ([T] => Set[T] => Option[T]) = [T] => (s: Set[T]) => s.headOption
val choose: PolyFunction{apply: [T](x$1: Set[T]): Option[T]} = <function1>

scala> val sets = Set(1) *: Set("foo") *: Tuple()
val sets: (Set[Int], Set[String]) = (Set(1),Set(foo))

scala> sets.map(choose)
1 |sets.map(choose)
  |         ^^^^^^
  |     Found:    (choose : PolyFunction{apply: [T](x$1: Set[T]): Option[T]})
  |     Required: PolyFunction{apply: [t](x$1: t): Any}

回答

def map[F[_]](f: [t] => (x$1: t) => F[t]): Map[Tuple, F]

type Choose[T] = T match {
  case Set[a] => Option[a]
}

def choose[T](t: T): Choose[T] = t match
  case set: Set[a] => set.headOption

// messy, but it works
@main def run =
  val sets = Set(1) *: Set("foo") *: Tuple()
  println(sets.map([T] => (t: T) => choose(t)))