Scala宏：在Scala中创建一个类的字段

Question

假设我有很多类似的数据类。下面是其被定义为如下的示例类User：

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

    ... 

    def foo = "bar" 

    ... 
} 

我对自动创建（在编译时）的方法，其返回的方式一个Map每个字段名称被映射到它的值，当它在运行时被调用。对于上面的例子，让我们说，我的方法被称为toMap：

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

myUser.toMap 

应该返回

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

你将如何与宏做到这一点？

这里是我做了什么：首先，我创建了一个Model类作为我所有的数据类的父类，并在那里实现的方法是这样的：

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

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

然后我在一个定义的宏实现独立Macros对象：

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)) 
    } 
} 

然而，当我尝试编译它，我得到SBT此错误：

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

Macros.scala正在编译中。这里是我的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) 

我在做什么错？我认为编译器在创建表达式时也会尝试评估字段标识符，但我不知道如何在表达式中正确返回它们。你能告诉我该怎么做吗？

非常感谢。

Answer 1

注意，这可以更优雅的完成，无需toString/c.parse业务：

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)) 
    } 
} 

还要注意的是你需要的c.resetAllAttrs位，如果你希望能够写出如下：

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

没有它你会在这种情况下得到一个令人困惑的ClassCastException。

顺便说一下，这里有一个技巧，我已经用来避免例如额外的类型参数。 user.toMap[User]书写时宏是这样的：

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：

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

，并不需要指定，我们正在谈论的是User。

Answer 2

有一个excellent blog post映射到/从案例类转换使用宏。