是否有一个类型类可以检查至少一个隐式类型的存在？

我有一个特质Foo[T, U]和一个类型级算法，给出L <: HList和目标类型U，告诉我是否存在T in L这样就存在一个隐含的Foo[T, U]在适用范围。这是使用以下类型类实现的：

trait Search[L <: HList, U]

object Search {
  def apply[L <: HList, U](implicit s: Search[L, U]): U = null

  ...
}

我们有以下内容：

object Test {
  type L = Int :: String :: HNil

  implicit val foo: Foo[String, Boolean] = null

  Search[L, Boolean] //compiles

  Search[L, Double] //does not compile
}

我希望的是，如果没有，则根本不会进行搜索Foo[T, U]对于任何T完全在范围内，因为那时我们已经知道该算法将无法完成。换句话说，我想要一个类型类trait Exists[F[_]]对于哪些实例存在当且仅当存在至少一个隐式F在范围内，所以函数Search.apply相反有签名：

def apply[L <: HList, U](implicit ev: Exists[Foo[_, U]], s: Search[L, U]): U = null

在这种情况下，编译器只会尝试解析s如果有任何隐含的Foo在适用范围。

这样的类型类可以定义吗？已经存在吗？

Try

import scala.language.experimental.macros
import scala.reflect.macros.{blackbox, contexts}

trait Exists[A]

object Exists {
  implicit def materialize[A]: Exists[A] = macro impl[A]

  def impl[A: c.WeakTypeTag](c: blackbox.Context): c.Tree = {
    import c.universe._
    val context = c.asInstanceOf[contexts.Context]
    val global: context.universe.type = context.universe
    val analyzer: global.analyzer.type = global.analyzer
    val callsiteContext = context.callsiteTyper.context

    val tpA = weakTypeOf[A]

    val searchResult = analyzer.inferImplicit(
      tree = EmptyTree.asInstanceOf[global.Tree],
      pt = tpA.asInstanceOf[global.Type],
      reportAmbiguous = false,
      isView = false,
      context = callsiteContext,
      saveAmbiguousDivergent = true,
      pos = c.enclosingPosition.asInstanceOf[global.Position]
    )

    val isAmbiguous = callsiteContext.reporter.firstError match {
      case Some(analyzer.AmbiguousImplicitTypeError(_,_)) => true
      case _ => false
    }

    if (searchResult.isSuccess || searchResult.isAmbiguousFailure || isAmbiguous) 
      q"new Exists[$tpA] {}"
    else c.abort(c.enclosingPosition, s"no implicit $tpA")    
  }
}

Test

// no implicit Int
// implicitly[Exists[Int]] // doesn't compile

implicit val i: Int = 1 
implicitly[Exists[Int]] // compiles

implicit val i: Int = 1 
implicit val i1: Int = 2 
implicitly[Exists[Int]] // compiles

我猜是原创的Search was

trait Foo[U, V]

trait Search[L <: HList, V]

trait LowPrioritySearch {
  implicit def tail[H, T <: HList, V](implicit search: Search[T, V]): Search[H :: T, V] = null
}

object Search extends LowPrioritySearch {
  def apply[L <: HList, U](implicit s: Search[L, U]): U = null.asInstanceOf[U]

  implicit def head[U, T <: HList, V](implicit foo: Foo[U, V]): Search[U :: T, V] = null
}

现在与Exists

def apply[L <: HList, U](implicit ev: Exists[Foo[_, U]], s: Search[L, U]): U = null.asInstanceOf[U]

也有效

Search[L, Boolean] //compiles
// Search[L, Double] //does not compile

在2.13.0中测试

libraryDependencies ++= Seq(
  scalaOrganization.value % "scala-reflect" % scalaVersion.value,
  scalaOrganization.value % "scala-compiler" % scalaVersion.value
)

Exists2.13.10 仍在工作。

Scala 3 中有一个 API 可以实现这一点

import scala.quoted.{Type, Quotes, Expr, quotes}

trait Exists[A]

object Exists {
  inline given [A]: Exists[A] = ${impl[A]}

  def impl[A: Type](using Quotes): Expr[Exists[A]] = {
    import quotes.reflect.*

    Implicits.search(TypeRepr.of[A]) match {
      case _: (ImplicitSearchSuccess | AmbiguousImplicits) => '{new Exists[A]{}}
      case _ => report.errorAndAbort(s"no implicit ${Type.show[A]}")
    }
  }
}

// no implicit Int
// summon[Exists[Int]] // doesn't compile

given Int = 1
summon[Exists[Int]] // compiles

given Int = 1
given i: Int = 2
summon[Exists[Int]] // compiles

斯卡拉 3.2.0。