scala - scala-meta: 引用符を外すときの型の不一致。見つかった: オプション[scala.meta.Type.Arg] 必須: scala.meta.Type

Question

scala-metaの新しいスタイルのマクロ注釈で少し遊んでいます。そこで、@Main アノテーションの例を拡張しました。

SConsumer.scala:

import scala.meta._


class SConsumer extends scala.annotation.StaticAnnotation {

  inline def apply(defn: Any): Any = meta {
    defn match {
      case q"case class $name($param: $tpe) { ..$stats }" =>
        val accept = q"def accept($param: $tpe): Unit = { ..$stats }"
        q"case class $name extends SConsumerProperty[${tpe}] { $accept }"
      case _ =>
        abort("error!")
    }
  }
}

SConsumerProperty.scala:

trait SConsumerProperty[T] {
  def accept(param: T): Unit
}

次のコンパイラ エラーが発生します。

Java HotSpot(TM) 64-Bit Server VM warning: ignoring option MaxPermSize=256M; support was removed in 8.0
[info] Loading project definition from /home/erik/Entwicklung/IntelliJ/Android/Apps/scalandroid/project
[info] Set current project to scalandroid (in build file:/home/erik/Entwicklung/IntelliJ/Android/Apps/scalandroid/)
[info] Packaging /home/erik/Entwicklung/IntelliJ/Android/Apps/scalandroid/target/scala-2.11/scalandroid_2.11-0.0.23-sources.jar ...
[info] Done packaging.
[info] Wrote /home/erik/Entwicklung/IntelliJ/Android/Apps/scalandroid/target/scala-2.11/scalandroid_2.11-0.0.23.pom
[info] Compiling 4 Scala sources to /home/erik/Entwicklung/IntelliJ/Android/Apps/scalandroid/target/android/intermediates/classes...
[error] /home/erik/Entwicklung/IntelliJ/Android/Apps/scalandroid/src/main/scala/com/bertderbecker/scalandroid/event/SConsumer.scala:14: type mismatch when unquoting;
[error]  found   : Option[scala.meta.Type.Arg]
[error]  required: scala.meta.Type
[error]         q"case class $name extends com.bertderbecker.scalandroid.event.SConsumerProperty[${tpe}] { $accept }"
[error]                                                                                        
[error] one error found
[error] (compile:compileIncremental) Compilation failed
[error] Total time: 6 s, completed 17.02.2017 16:16:02
Process finished with exit code 1

では、Type.Arg を Type に変換する方法は?