Question

我正在学习scala，要分配作业，我必须写一个宏。
宏应浏览表达式树，然后构建自定义Expression。通过添加println(showRaw(exprTree))，我设法“看了一下”表达式。但是，我仍然无法对其进行迭代并构建Expression

我有以下两个文件：
ExpressionImplicits.scala ：

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

// Expression is defined elsewhere and mainly only overrides toString()
abstract class Expression
case class Var(name: String) extends Expression
case class Number(num: Double) extends Expression
case class BinOp(operator: String, left: Expression, right: Expression) extends Expression

class ExpressionImplicitsImpl(val c: Context) {

  import c.universe._

  // Task complete macro
  // Add necessary definitions here
  // This definition was added by me
  def expr(exprTree: c.Expr[AnyRef]): c.Expr[Expression] = {
    println(showRaw(exprTree)) 
    //prints
    //Expr(Function(List(ValDef(Modifiers(PARAM), TermName("x"), TypeTree().setOriginal(Select(Ident(scala), scala.Double)), EmptyTree)), Apply(Select(Apply(Select(Ident(TermName("x")), TermName("$times")), List(Literal(Constant(2)))), TermName("$plus")), List(Apply(Select(Literal(Constant(3.0)), TermName("$times")), List(Ident(TermName("x"))))))))
    //Expr(Function(List(ValDef(Modifiers(PARAM), TermName("x"), TypeTree().setOriginal(Select(Ident(scala), scala.Double)), EmptyTree), ValDef(Modifiers(PARAM), TermName("y"), TypeTree().setOriginal(Select(Ident(scala), scala.Double)), EmptyTree)), Apply(Select(Apply(Select(Ident(TermName("x")), TermName("$times")), List(Ident(TermName("y")))), TermName("$times")), List(Ident(TermName("x"))))))

  }
}

// This definition is given
object ExpressionImplicits {
  def expr(exprTree: AnyRef): Expression = macro ExpressionImplicitsImpl.expr
}

ExpressionsTest.scala ：

object ExpressionsTest {
  def main(args: Array[String]) {

    import ExpressionImplicits._
    val e1 = expr { (x: Double) => (x * 2) + (3.0 * x) }

    println(e1) // BinOp(+,BinOp(*,Var(x),Number(2.0)),BinOp(*,Number(3.0),Var(x)))

    val e2 = expr { (x: Double, y: Double) => x * y * x }
    println(e2) // BinOp(*,BinOp(*,Var(x),Var(y)),Var(x))

    // val e3 = expr { (x: Double) => x.toInt } // Fails during compilation
  }

}

Answer 1

您非常亲密。现在，您只需要匹配showRaw转储的表达式。

这是完整的solution：

object ExpressionImplicits {

  def expr(expr: AnyRef): Expression = macro expr_impl

  def expr_impl(c: blackbox.Context)(expr: c.Expr[AnyRef]): c.Expr[Expression] = {
    import c.universe._

    def treeToExpression(functionBody: c.Tree): c.Expr[Expression] = {
      functionBody match {
        case Apply(Select(leftTree, operator), List(rightTree)) =>
          val operatorName = Constant(operator.toString)
          c.Expr[Expression](q"sk.ygor.stackoverflow.q53326545.macros.BinOp($operatorName, ${treeToExpression(leftTree)}, ${treeToExpression(rightTree)})")
        case Ident(TermName(varName)) =>
          c.Expr[Expression](q"sk.ygor.stackoverflow.q53326545.macros.Var($varName)")
        case Literal(Constant(num)) if num.isInstanceOf[java.lang.Number] =>
          c.Expr[Expression](q"sk.ygor.stackoverflow.q53326545.macros.Number(${num.asInstanceOf[java.lang.Number].doubleValue()})")
        case unsupported =>
          sys.error("Unsupported function body: " + unsupported);
      }
    }

    expr.tree match {
      case Function(_, body) => treeToExpression(body)
      case unsupported =>
        sys.error("Only functions are accepted. Got: " + unsupported);
    }

  }
}

您应该尝试了解发生了什么：

我们正在通过模式匹配和递归进行树遍历。链接已在评论中提供：https://docs.scala-lang.org/overviews/reflection/symbols-trees-types.html#traversing-trees
第一个匹配项仅检查，顶层树是函数的定义
我们在功能体上进行递归匹配
- List(rightTree)的意思是，我们期望方法仅包含一个参数，例如x.foo(y)，x foo y，x.+(y)，x + y，但不是x.foo()，x.foo(y, z)，x.+(y, z)
我们使用Scala宏quasiqotes构建并合并输出树的各个部分
我们为BinOp，Var和Number使用完全限定的名称，这样宏的使用者就不必导入这些子类了。

探索Scala宏中的表达式树

1 个答案: