Question

我正在使用macro annotations的新风格scala-meta。所以我扩展了@Main注释的例子：

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
}

它给出了以下编译器错误：

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

那么，如何将Type.Arg转换为Type？

Answer 1

我自己被这个咬了。一般来说，对于这样的情况，咨询Tree sources是件好事。所有Type均为Type.Arg（请参阅trait Type extends X with Type.Arg，但有两个Type.Arg不是Type：Type.Arg.ByName和Type.Arg.Repeated。

例如，

q"def foo(k: Int, a: => Int, b: String*)".structure
res22: String = """
Decl.Def(Nil, Term.Name("foo"), Nil,
         Seq(Seq(Term.Param(Nil, Term.Name("k"), Some(Type.Name("Int")), None),
                 Term.Param(Nil, Term.Name("a"), Some(Type.Arg.ByName(Type.Name("Int"))), None),
                 Term.Param(Nil, Term.Name("b"), Some(Type.Arg.Repeated(Type.Name("String"))), None))),
         Type.Name("Unit"))
"""

我们可以创建一个辅助工具来将Type.Arg转换为Type

def toType(arg: Type.Arg): Type = arg match {
  case Type.Arg.Repeated(tpe) => tpe
  case Type.Arg.ByName(tpe) => tpe
  case tpe: Type => tpe
}

scala-meta：不引用时的类型不匹配; found：需要选项[scala.meta.Type.Arg]：scala.meta.Type

1 个答案: