8
votes

The following is an obvious variadic function:

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

We can define a macro in a similar way:

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

fun(1,"1",1.0)

But in this case, all the arguments are typed as Any. In fact, the compiler knows the types at compile-time, but hides it from us. Is it possible to get a list of arguments and their types in a macro?

1

1 Answers

6
votes

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.

Note also that this won't work if the argument is a sequence with the _* type ascription, of course, and that you'll need to write something like the following if you want to catch this case and provide a useful error message:

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!")
      }
  }
}

See my question here and bug report here for more discussion.