是否可以使用可变参数定义宏,并为每个参数获取一个类型？

Question

以下是一个明显的可变函数:

def fun(xs: Any*) = ???

我们可以用类似的方式定义一个宏:

def funImpl(c: Context)(xs: c.Expr[Any]*) = ???

fun(1,"1",1.0)

但在这种情况下,所有参数都被输入为Any.实际上,编译器在编译时知道类型,但是将它隐藏起来.是否有可能得到的参数列表,并在宏它们的类型？

Answer 1

Sure—for example:

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

object Demo {
  def at(xs: Any*)(i: Int) = macro at_impl
  def at_impl(c: Context)(xs: c.Expr[Any]*)(i: c.Expr[Int]) = {
    import c.universe._

    // First let's show that we can recover the types:
    println(xs.map(_.actualType))

    i.tree match {
      case Literal(Constant(index: Int)) => xs.lift(index).getOrElse(
        c.abort(c.enclosingPosition, "Invalid index!")
      )
      case _ => c.abort(c.enclosingPosition, "Need a literal index!")
    }
  }
}

And then:

scala> Demo.at(1, 'b, "c", 'd')(1)
List(Int(1), Symbol, String("c"), Char('d'))
res0: Symbol = 'b

scala> Demo.at(1, 'b, "c", 'd')(2)
List(Int(1), Symbol, String("c"), Char('d'))
res1: String = c

Note that the inferred types are precise and correct.

另请注意,如果参数是具有_*类型ascription 的序列,当然,如果要捕获此情况并提供有用的错误消息,则需要编写类似以下内容的内容:

def at_impl(c: Context)(xs: c.Expr[Any]*)(i: c.Expr[Int]) = {
  import c.universe._

  xs.toList.map(_.tree) match {
    case Typed(_, Ident(tpnme.WILDCARD_STAR)) :: Nil => 
      c.abort(c.enclosingPosition, "Needs real varargs!")
    case _ =>
      i.tree match {
        case Literal(Constant(index: Int)) => xs.lift(index).getOrElse(
          c.abort(c.enclosingPosition, "Invalid index!")
        )
        case _ => c.abort(c.enclosingPosition, "Need a literal index!")
      }
  }
}

见我的问题在这里和bug报告,这里更多的讨论.