Getting a structural type with an anonymous class's methods from a macro

Solution 1:

This question is answered in duplicate by Travis here. There are links to the issue in the tracker and to Eugene's discussion (in the comments and mailing list).

In the famous "Skylla and Charybdis" section of the type checker, our hero decides what shall escape dark anonymity and see the light as a member of the structural type.

There are a couple of ways to trick the type checker (which do not entail Odysseus's ploy of hugging a sheep). The simplest is to insert a dummy statement so that the block doesn't look like an anonymous class followed by its instantiation.

If the typer notices that you're a public term that isn't referenced by the outside, it will make you private.

object Mac {
  import scala.language.experimental.macros
  import scala.reflect.macros.Context

  /* Make an instance of a structural type with the named member. */
  def bar(name: String): Any = macro bar_impl

  def bar_impl(c: Context)(name: c.Expr[String]) = {
    import c.universe._
    val anon = TypeName(c.freshName)
    // next week, val q"${s: String}" = name.tree
    val Literal(Constant(s: String)) = name.tree
    val A    = TermName(s)
    val dmmy = TermName(c.freshName)
    val tree = q"""
      class $anon {
        def $A(i: Int): Int = 2 * i
      }
      val $dmmy = 0
      new $anon
    """
      // other ploys
      //(new $anon).asInstanceOf[{ def $A(i: Int): Int }]
      // reference the member
      //val res = new $anon
      //val $dmmy = res.$A _
      //res
      // the canonical ploy
      //new $anon { }  // braces required
    c.Expr(tree)
  }
}