Scala Macros:使用Scala中的类字段制作地图

Question

假设我有很多类似的数据类.这是一个示例类User,定义如下:

case class User (name: String, age: Int, posts: List[String]) {
  val numPosts: Int = posts.length

  ...

  def foo = "bar"

  ...
}

我感兴趣的是自动创建一个方法(在编译时),Map该方法以在运行时调用每个字段名称时将其映射到其值的方式返回.对于上面的例子,让我们说我的方法被调用toMap:

val myUser = User("Foo", 25, List("Lorem", "Ipsum"))

myUser.toMap

应该回来

Map("name" -> "Foo", "age" -> 25, "posts" -> List("Lorem", "Ipsum"), "numPosts" -> 2)

你会如何用宏来做到这一点？

这就是我所做的:首先,我创建了一个Model类作为我所有数据类的超类,并在那里实现了这样的方法:

abstract class Model {
  def toMap[T]: Map[String, Any] = macro toMap_impl[T]
}

class User(...) extends Model {
  ...
}

然后我在一个单独的Macros对象中定义了一个宏实现:

object Macros {
  import scala.language.experimental.macros
  import scala.reflect.macros.Context
  def getMap_impl[T: c.WeakTypeTag](c: Context): c.Expr[Map[String, Any]] = {
    import c.universe._

    val tpe = weakTypeOf[T]

    // Filter members that start with "value", which are val fields
    val members = tpe.members.toList.filter(m => !m.isMethod && m.toString.startsWith("value"))

    // Create ("fieldName", field) tuples to construct a map from field names to fields themselves
    val tuples =
      for {
        m <- members
        val fieldString = Literal(Constant(m.toString.replace("value ", "")))
        val field = Ident(m)
      } yield (fieldString, field)

    val mappings = tuples.toMap

    /* Parse the string version of the map [i.e. Map("posts" -> (posts), "age" -> (age), "name" -> (name))] to get the AST
     * for the map, which is generated as:
     * 
     * Apply(Ident(newTermName("Map")), 
     *   List(
     *     Apply(Select(Literal(Constant("posts")), newTermName("$minus$greater")), List(Ident(newTermName("posts")))), 
     *     Apply(Select(Literal(Constant("age")), newTermName("$minus$greater")), List(Ident(newTermName("age")))), 
     *     Apply(Select(Literal(Constant("name")), newTermName("$minus$greater")), List(Ident(newTermName("name"))))
     *   )
     * )
     * 
     * which is equivalent to Map("posts".$minus$greater(posts), "age".$minus$greater(age), "name".$minus$greater(name)) 
     */
    c.Expr[Map[String, Any]](c.parse(mappings.toString))
  }
}

但是当我尝试编译它时,我从sbt得到了这个错误:

[error] /Users/emre/workspace/DynamoReflection/core/src/main/scala/dynamo/Main.scala:9: not found: value posts
[error]     foo.getMap[User]
[error]               ^

首先编译Macros.scala.这是我的Build.scala的片段:

lazy val root: Project = Project(
    "root",
    file("core"),
    settings = buildSettings
  ) aggregate(macros, core)

  lazy val macros: Project = Project(
    "macros",
    file("macros"),
    settings = buildSettings ++ Seq(
      libraryDependencies <+= (scalaVersion)("org.scala-lang" % "scala-reflect" % _))
  )

  lazy val core: Project = Project(
    "core",
    file("core"),
    settings = buildSettings
  ) dependsOn(macros)

我究竟做错了什么？我认为编译器在创建表达式时也会尝试计算字段标识符,但我不知道如何在表达式中正确返回它们.你能告诉我怎么做吗？

首先十分感谢.

Answer 1

请注意,如果没有toString/ c.parsebusiness ,这可以更优雅地完成:

import scala.language.experimental.macros

abstract class Model {
  def toMap[T]: Map[String, Any] = macro Macros.toMap_impl[T]
}

object Macros {
  import scala.reflect.macros.Context

  def toMap_impl[T: c.WeakTypeTag](c: Context) = {
    import c.universe._

    val mapApply = Select(reify(Map).tree, newTermName("apply"))

    val pairs = weakTypeOf[T].declarations.collect {
      case m: MethodSymbol if m.isCaseAccessor =>
        val name = c.literal(m.name.decoded)
        val value = c.Expr(Select(c.resetAllAttrs(c.prefix.tree), m.name))
        reify(name.splice -> value.splice).tree
    }

    c.Expr[Map[String, Any]](Apply(mapApply, pairs.toList))
  }
}

另请注意,c.resetAllAttrs如果您希望能够编写以下内容,则需要该位:

User("a", 1, Nil).toMap[User]

如果没有它,你会ClassCastException在这种情况下感到困惑.

顺便说一句,这是我用来避免额外类型参数的技巧,例如user.toMap[User]在写这样的宏时:

import scala.language.experimental.macros

trait Model

object Model {
  implicit class Mappable[M <: Model](val model: M) extends AnyVal {
    def asMap: Map[String, Any] = macro Macros.asMap_impl[M]
  }

  private object Macros {
    import scala.reflect.macros.Context

    def asMap_impl[T: c.WeakTypeTag](c: Context) = {
      import c.universe._

      val mapApply = Select(reify(Map).tree, newTermName("apply"))
      val model = Select(c.prefix.tree, newTermName("model"))

      val pairs = weakTypeOf[T].declarations.collect {
        case m: MethodSymbol if m.isCaseAccessor =>
          val name = c.literal(m.name.decoded)
          val value = c.Expr(Select(model, m.name))
          reify(name.splice -> value.splice).tree
      }

      c.Expr[Map[String, Any]](Apply(mapApply, pairs.toList))
    }
  }
}

现在我们可以写下面的内容:

scala> println(User("a", 1, Nil).asMap)
Map(name -> a, age -> 1, posts -> List())

并且不需要指定我们正在讨论的问题User.

Answer 2

在使用宏的案例类转换的地图上有一篇很棒的博客文章.