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

[Aurangzeb]

Suppose we want to write a macro that defines an anonymous class with some type members or methods, and then creates an instance of that class that's statically typed as a structural type with those methods, etc. This is possible with the macro system in 2.10.0, and the type member part is extremely easy:

object MacroExample extends ReflectionUtils {
  import scala.language.experimental.macros
  import scala.reflect.macros.Context
 
  def foo(name: String): Any = macro foo_impl
  def foo_impl(c: Context)(name: c.Expr[String]) = {
    import c.universe._
 
    val Literal(Constant(lit: String)) = name.tree
    val anon = newTypeName(c.fresh)
 
    c.Expr(Block(
      ClassDef(
        Modifiers(Flag.FINAL), anon, Nil, Template(
          Nil, emptyValDef, List(
            constructor(c.universe),
            TypeDef(Modifiers(), newTypeName(lit), Nil, TypeTree(typeOf[Int]))
          )
        )
      ),
      Apply(Select(New(Ident(anon)), nme.CONSTRUCTOR), Nil)
    ))
  }
}

(Where ReflectionUtils is a convenience trait that provides my constructor method.)

This macro lets us specify the name of the anonymous class's type member as a string literal:

scala> MacroExample.foo("T")
res0: AnyRef{type T = Int} = $1$$1@7da533f6

Note that it's appropriately typed. We can confirm that everything's working as expected:

scala> implicitly[res0.T =:= Int]
res1: =:=[res0.T,Int] = <function1>
 

Now suppose that we try to do the same thing with a method:

def bar(name: String): Any = macro bar_impl
def bar_impl(c: Context)(name: c.Expr[String]) = {
  import c.universe._
 
  val Literal(Constant(lit: String)) = name.tree
  val anon = newTypeName(c.fresh)
 
  c.Expr(Block(
    ClassDef(
      Modifiers(Flag.FINAL), anon, Nil, Template(
        Nil, emptyValDef, List(
          constructor(c.universe),
          DefDef(
            Modifiers(), newTermName(lit), Nil, Nil, TypeTree(),
            c.literal(42).tree
          )
        )
      )
    ),
    Apply(Select(New(Ident(anon)), nme.CONSTRUCTOR), Nil)
  ))
}

But when we try it out, we don't get a structural type

scala> MacroExample.bar("test")
res1: AnyRef = $1$$1@da12492

But if we stick an extra anonymous class in there:

def baz(name: String): Any = macro baz_impl
def baz_impl(c: Context)(name: c.Expr[String]) = {
  import c.universe._
 
  val Literal(Constant(lit: String)) = name.tree
  val anon = newTypeName(c.fresh)
  val wrapper = newTypeName(c.fresh)
 
  c.Expr(Block(
    ClassDef(
      Modifiers(), anon, Nil, Template(
        Nil, emptyValDef, List(
          constructor(c.universe),
          DefDef(
            Modifiers(), newTermName(lit), Nil, Nil, TypeTree(),
            c.literal(42).tree
          )
        )
      )
    ),
    ClassDef(
      Modifiers(Flag.FINAL), wrapper, Nil,
      Template(Ident(anon) :: Nil, emptyValDef, constructor(c.universe) :: Nil)
    ),
    Apply(Select(New(Ident(wrapper)), nme.CONSTRUCTOR), Nil)
  ))
}

It works:

scala> MacroExample.baz("test")
res0: AnyRef{def test: Int} = $2$$1@6663f834
 
scala> res0.test
res1: Int = 42

This is extremely handy—it lets you do things like this, for example—but I don't understand why it works, and the type member version works, but not bar. I know this may not be defined behavior, but does it make any sense? Is there an cleaner way to get a structural type (with the methods on it) from a macro?