usage.scala

Question

我有以下代码：

usage.scala

object Test extends App {
   import Macros._

   val f = 1
   Macros.get(f) 
}

macros.scala

import language.experimental.macros
import scala.reflect.macros.Context

object Macros {
  def get(a: Int) = macro getImpl

  def getImpl(c: Context)(a: c.Expr[Int]) = {
    import c.universe._

     println(showRaw(a))
  }
}

它返回：

Expr(Select(This(newTypeName("Test")), newTermName("f")))

如何从termName("f") 1值中提取？有可能用宏吗？

Answer 1

总的来说 - 没有。 f可以是抽象的或定义为外部呼叫或用户输入或随机或许多其他情况之一。

但在某些特殊情况下你可以得到它。你知道几乎所有编译器都知道了！

看看c.enclosingClass：

object Macros {
  def get(a: Int) = macro getImpl

  def getImpl(c: Context)(a: c.Expr[Int]) = {
    import c.universe._
    println(showRaw(c.enclosingClass))
    c.Expr[Unit](Literal(Constant(())))
  }
}    

object Test { val f = 1; Macros.get(f) }
// ModuleDef(Modifiers(), $line55.$read.$iw.$iw.$iw.$iw.Test, Template(List(Select(Ident(scala), newTypeName("AnyRef"))), emptyValDef, List(DefDef(Modifiers(), nme.CONSTRUCTOR, List(), List(List()), TypeTree(), Block(List(Apply(Select(Super(This(newTypeName("Test")), tpnme.EMPTY), nme.CONSTRUCTOR), List())), Literal(Constant(())))), ValDef(Modifiers(), newTermName("f "), TypeTree(), Literal(Constant(1))), Apply(Select(Ident(newTermName("Macros")), newTermName("get")), List(Ident(newTermName("f")))))))

这里有趣的部分是ValDef(Modifiers(), newTermName("f "), TypeTree(), Literal(Constant(1)))。

我们必须提取它：

object Macros {
  def get(a: Int) = macro getImpl

  def getImpl(c: Context)(a: c.Expr[Int]) = {
    import c.universe._

    val (enclosing, name) = a.tree match {
      case Select(This(enclosing), name) => enclosing -> name
      case _ => c.abort(c.enclosingPosition, "Not a `this` memver")
    }
    val impl = c.enclosingClass match {
      case impl: ImplDef if impl.name.toString == enclosing.toString => impl
      case impl: ImplDef => c.abort(c.enclosingPosition, "Should search in another parent")
      case _ => c.abort(c.enclosingPosition, "Not an `ImplDef`")
    }
    val body = impl.children.collect{
      case Template(_, _, body) => body
    } match {
      case Seq(body) => body
      case _ => c.abort(c.enclosingPosition, "Should be a single template.")
    }
    val rhss = body.collect{
      case ValDef(_, valName, _, rhs) if valName.toString == name.toString => rhs
    }
    val rhs = rhss match {
      case Seq(rhs) => rhs
      case Seq() => c.abort(c.enclosingPosition, "Not found. Maybe it's a DefDef or somethong else")
      case _ => c.abort(c.enclosingPosition, "Some other error.")
    }
    val res = rhs match {
      case Literal(Constant(i: Int)) => i
      case Literal(Constant(_)) => c.abort(c.enclosingPosition, "Literal, but not an Int.")
      case _ => c.abort(c.enclosingPosition, "Implemented not as literal.")
    }
    println(s"Int value in this particular case: $res")
    c.Expr[Any](Literal(Constant(res)))
  }
}

结果：

object Test { val f = 1; Macros.get(f) }
// Int value in this particular case: 1

因此我们在编译时有一个f的值。

我很确定这不是你所期望的。

Scala Macro获得术语名称的值

usage.scala

macros.scala

1 个答案: