是否可以在Scala中使用名称和命名参数的值来创建Map [String,Any]？

Question

我正在编写REST Web服务的包装器,我想要强类型的Scala API.

以下是我到目前为止所做的事情:

def getMentions(count: Option[Int] = None,
                sinceID: Option[TweetID] = None,
                maxID: Option[TweetID] = None,
                trimUser: Option[Boolean] = None,
                contributorDetails: Option[Boolean] = None,
                includeEntities: Option[Boolean] = None) : List[Tweet] = {
val parameters = Map("count" -> count,
                     "since_id" -> sinceID,
                     "max_id" -> maxID,
                     "trim_user" -> trimUser,
                     "contributor_details" -> contributorDetails,
                     "include_entities" -> includeEntities)
/* 
 * Convert parameters, which is a Map[String,Any] to a Map[String,String]
 * (Removing Nones) and pass it to an object in charge of generating the request.
 */
...
}

这种方法很有效,但它需要我手动生成parameters地图.如果我能够访问代表参数及其值的Map,那么我所做的将更加清晰.

Answer 1

你可以通过运行时反射来做到这一点,我相信你会得到答案,告诉你如何,如果你想要,但这实际上是Scala 2.10的宏的一个简洁的用例,所以这里.首先假设我们有一个名为的文件ParamMapMaker.scala:

object ParamMapMaker {
  def paramMap: Map[String, Any] = macro paramMapImpl

  def paramMapImpl(c: scala.reflect.macros.Context) = {
    import c.universe._

    val params = c.enclosingMethod match {
      case DefDef(_, _, _, ps :: Nil, _, _) =>
        ps.map(p =>
          reify((
            c.Expr[String](Literal(Constant(p.name.decoded))).splice,
            c.Expr[Any](Ident(p.symbol)).splice
          )).tree
        )
      case _ => c.abort(c.enclosingPosition, "Can't call paramMap here!")
    }

    c.Expr[Map[String, Any]](Apply(Select(Ident("Map"), "apply"), params))
  }
}

我会留下蛇套管地图键作为读者的(简单)练习.

我们还有一个测试文件(命名Test.scala):

object Test extends App {
  def foo(hello: String, answer: Int) = ParamMapMaker.paramMap

  println(foo("world", 42))
}

现在我们编译这两个:

scalac -language:experimental.macros ParamMapMaker.scala
scalac Test.scala

当我们运行时,Test我们将得到以下内容:

Map(hello -> world, answer -> 42)

关于这一点的好处是没有运行时反射的开销.如果我们使用编译测试文件-Ymacro-debug-verbose,我们会看到foo在编译时为生成器生成了以下代码(实际上):

Map.apply[String, Any](
  scala.Tuple2.apply[String, String]("hello", hello),
  scala.Tuple2.apply[String, Int]("answer", answer)
)

正如我们所期望的那样.