Scala3 宏召唤 TypeTree 的类型类实例（无类型参数）

Question

trait Show[T] {
  def show(t: T): String
}

给这样的Show类型class，我想为案例class生成节目

 def caseClassShow[A](using Type[A], Quotes): Expr[Show[A]] = {
    import quotes.reflect._
    def shows(caseClassExpr: Expr[A]): Expr[String] = { 
      val caseClassTerm = caseClassExpr.asTerm
      val parts = TypeRepr.of[A].typeSymbol.caseFields.collect {
        case cf if cf.isValDef =>
          val valDefTree = cf.tree.asInstanceOf[ValDef]
          val valType = valDefTree.tpt
          val showCtor = TypeTree.of[Show[_]]
          val valShowType = Applied(showCtor, List(valType))
          val showInstance = Expr.summon[valShowType] // compile error, how to summon the instance here
          val valuePart = Apply(Select.unique(showInstance, "show"), List(Select(caseClassTerm, cf)))
          '{s"${Expr(cf.name)}:${valuePart}"}
      }
      val strParts = Expr.ofList(parts)
      '{$strParts.mkString(",")}
    }
    '{
      new Show[A] {
        def show(a: A) = {
          ${shows('{a})}
        }
      }
    }
  }

但是 showInstance 部分不会编译，那么如何在这里调用隐式 Show[X] 呢？

Answer 1

Implicits.search 如果 Expr.summon

没有可用的类型参数，则可用于调用隐式实例

  val valDefTree = cf.tree.asInstanceOf[ValDef]
  val valType = valDefTree.tpt
  val showCtor = TypeRepr.typeConstructorOf(classOf[Show[_]])
  val valShowType = showCtor.appliedTo(valType.tpe)
  Implicits.search(valShowType) match {
    case si: ImplicitSearchSuccess =>
      val siExpr: Expr[Show[Any]] = si.tree.asExpr.asInstanceOf[Expr[Show[Any]]]
      val valueExpr = Select(caseClassTerm, cf).asExpr
      '{$siExpr.show($valueExpr)}
  }