Get the module symbol, given I have the module class, scala macro

Question

I\'m trying to build a simple typeclass IsEnum[T] using a macro.

I use knownDirectSubclasses to get all the direct subclasses if T,

Answer 1

In 2.13 you can materialize scala.ValueOf

val instanceTree = c.inferImplicitValue(appliedType(typeOf[ValueOf[_]].typeConstructor, subSymbol.asClass.toType))
q"$instanceTree.value"

Tree will be different

sealed trait A
object A {
  case object B extends A
  case object C extends A
}

//scalac: Seq(new scala.ValueOf(A.this.B).value, new scala.ValueOf(A.this.C).value)

but at runtime it's still Seq(B, C).

In 2.12 shapeless.Witness can be used instead of ValueOf

val instanceTree = c.inferImplicitValue(appliedType(typeOf[Witness.Aux[_]].typeConstructor, subSymbol.asClass.toType))
q"$instanceTree.value"

//scalac: Seq(Witness.mkWitness[App.A.B.type](A.this.B.asInstanceOf[App.A.B.type]).value, Witness.mkWitness[App.A.C.type](A.this.C.asInstanceOf[App.A.C.type]).value)

libraryDependencies += "com.chuusai" %% "shapeless" % "2.4.0-M1" // in 2.3.3 it doesn't work

---

In Shapeless they use kind of

subSymbol.asClass.toType match {
  case ref @ TypeRef(_, sym, _) if sym.isModuleClass => mkAttributedQualifier(ref)
}

https://github.com/milessabin/shapeless/blob/master/core/src/main/scala/shapeless/singletons.scala#L230

or in our case simply

mkAttributedQualifier(subSymbol.asClass.toType)

but their mkAttributedQualifier also uses downcasting to compiler internals and the tree obtained is like Seq(A.this.B, A.this.C).

Also

Ident(subSymbol.companionSymbol)

seems to work (tree is Seq(B, C)) but .companionSymbol is deprecated (in scaladocs it's written "may return unexpected results for module classes" i.e. for objects).

Following approach similar to the one used by @MateuszKubuszok in his library enumz you can try also

val objectName = symbol.fullName
c.typecheck(c.parse(s"$objectName"))

and the tree is Seq(App.A.B, App.A.C).

Finally, if you're interested in the tree Seq(B, C) (and not some more complicated tree) it seems you can replace

Ident(subSymbol.asInstanceOf[scala.reflect.internal.Symbols#Symbol].sourceModule.asInstanceOf[Symbol])

with more conventional

Ident(subSymbol.owner.info.decl(subSymbol.name.toTermName))