Unable to use self reference in refinement within quotes

[steinybot]

Compiler version

3.3.0-RC3

Minimized code

import scala.quoted.*

transparent inline def thing =
  ${ thingImpl }

def thingImpl(using Quotes): Expr[Any] =
  '{
    def makeThing: { def me: this.type } = ???
    makeThing
  }

Output

access to <refinement>.this from wrong staging level:
 - the definition is at level 0,
 - but the access is at level 1.

Expectation

I think this should compile.

I know that self reference in refinement is deprecated but I haven't figured out another way to refer to the same type, especially since cyclic type references are illegal which is annoying.

I can't figure out how to do this with the Quotes API either. Refinement doesn't take a symbol so there is nothing to give to This.

[nicolasstucki]

This should work.

We are probably not recording the level of the refinement or the this definition, in which case the default level of 0 is used.

[sjrd]

I can't figure out how to do this with the Quotes API either. Refinement doesn't take a symbol so there is nothing to give to This.

You need to wrap your refinement with a RecursiveType and use a RecursiveThis inside to get access to this.

[steinybot]

I can't figure out how to do this with the Quotes API either. Refinement doesn't take a symbol so there is nothing to give to This.

You need to wrap your refinement with a RecursiveType and use a RecursiveThis inside to get access to this.

I tried that but the mistake I made was that I was literally recursing it as in:

  val refinement = labels.foldLeft(RecursiveType(_ => TypeRepr.of[Object])) {
    case (result, label) =>
      val methodType = MethodType(Nil)(_ => Nil, _ => result.recThis)
      val resultWithMethod = Refinement(result, label, methodType)
      RecursiveType(_ => resultWithMethod)
  }

Which blows up the pickler with an AssertionError.

This does indeed work:

  val recursiveType = RecursiveType(_ => TypeRepr.of[Object])
  val refinement = labels.foldLeft[TypeRepr](recursiveType) {
    case (result, label) =>
      val methodType = MethodType(Nil)(_ => Nil, _ => recursiveType.recThis)
      Refinement(result, label, methodType)
  }

Thank you! I really appreciate the quick response and providing the answer I was actually looking for all along.

[steinybot]

Woops that's still not right. It is coming out as me: this instead of me: this.type.

Here is a more complete example:

import scala.quoted.*

class SelectMe extends Selectable {
  def selectDynamic(name: String): Any = name match
    case "name" => "Bob"
    case "me" => this
}

transparent inline def thing =
  ${ thingImpl }

def thingImpl(using Quotes): Expr[Any] =
  import quotes.reflect._

  val recursiveType = RecursiveType(_ => TypeRepr.of[Object])
  val refinementType =
    Refinement(
      Refinement(
        recursiveType,
        "name",
        TypeRepr.of[String]
      ),
      "me",
      recursiveType.recThis
    ).asType

  val result = refinementType match
    case '[t] => '{
      (new SelectMe).asInstanceOf[t]
    }
  println(result.show)
  result

def thing2 = new SelectMe().asInstanceOf[java.lang.Object {
  val name: java.lang.String
  val me: this.type
}]

This:

thing.me.name

Fails with:

[info] compiling 1 Scala source to /Users/jason/src/bug-reports/app/target/scala-3.3.0-RC4/classes ...
new SelectMe().asInstanceOf[java.lang.Object {
  val name: java.lang.String
  val me: this
}]
[error] -- [E008] Not Found Error: /Users/jason/src/bug-reports/app/src/main/scala/Main.scala:51:11
[error] 51 |  thing.me.name
[error]    |  ^^^^^^^^^^^^^
[error]    |value name is not a member of {z1 => Object} - did you mean {z1 => Object}.wait?

But thing2.me.name works.

[sjrd]

You can only use recThis "lexically" within the recursive type:

  val recursiveType = RecursiveType({ recType =>
    Refinement(
      Refinement(
        TypeRepr.of[Object],
        "name",
        TypeRepr.of[String]
      ),
      "me",
      recType.recThis
    )
  }).asType

[nicolasstucki]

This is already fixed.