Using type parameter in anonymous class in Scala

Question

I'm trying to declare attribute of parametrized type inside anonymous class. This works in Java, in Scala (2.9) however I get compile error:

Parameter type in structural refinement may not refer to an abstract type defined outside that refinement

This is the code:

object DemoFail extends App {
  def it[T <: AnyRef](x: T) = new Iterator[T] {
    var i = x // here is the error
    def next = i
    def hasNext = true
  }
  for (i ← it(int2Integer(4))) println(i)
}

I can get it to work by "erasing" types manually:

object DemoOK extends App {
  def it[T <: AnyRef](x: T) = new Iterator[T] {
    var i: AnyRef = x
    def next = i.asInstanceOf[T]
    def hasNext = true
  }
  for (i ← it(int2Integer(4))) println(i)
}

So the question is: why can't the compiler do it for me ?

By adding a public variable to your iterator, you create a structural type that is a subtype of Iterator. It'll work if you change i to a private variable. — Kris Nuttycombe
– Kris Nuttycombe, Commented May 29, 2011 at 21:59
Indeed, I don't know how could I miss such an obvious solution. — Marcin Wisnicki
– Marcin Wisnicki, Commented May 29, 2011 at 23:54
@Kris: please respond below with an answer so I can properly accept it. — Marcin Wisnicki
– Marcin Wisnicki, Commented May 29, 2011 at 23:55
@Kris So this means that any access to i would happen through Java reflection, right? — Jean-Philippe Pellet
– Jean-Philippe Pellet, Commented May 30, 2011 at 7:22
I think that internal accesses would not use reflection, but you'd have to look at the bytecode to be sure. — Kris Nuttycombe
– Kris Nuttycombe, Commented Jun 10, 2011 at 16:24

Kris Nuttycombe · Accepted Answer · 2011-06-10 16:23:26Z

1

By adding a public variable to your iterator, you create a structural type that is a subtype of Iterator. It'll work if you change i to a private variable.

answered Jun 10, 2011 at 16:23

Kris Nuttycombe

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Comments

Rex Kerr · Accepted Answer · 2011-05-29 21:48:23Z

1

I am, unfortunately, not sure why this doesn't work. But here is an alternate workaround that avoids casts:

def it[T <: AnyRef](x: T) = {
  class Forever extends Iterator[T] {
    var i = x
    def next = i
    def hasNext = true
  }
  new Forever
}

answered May 29, 2011 at 21:48

Rex Kerr

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1 Comment

Kris Nuttycombe Over a year ago

This works because it avoids the structural type I mentioned above.

Alex Archambault · Accepted Answer · 2014-06-06 14:51:40Z

A quick fix would be to avoid the structural return type:

object DemoFail extends App {
  // The same with an explicit (non structural) return type
  //                       vvvvvvvvvvvvv
  def it[T <: AnyRef](x: T): Iterator[T] = 
    new Iterator[T] {
      var i = x // no more error
      def next = i
      def hasNext = true
    }
  for (i ← it(int2Integer(4))) println(i)
}

Indeed, method it on object DemoFail does not have an explicit return type. Hence this return type is inferred by the compiler.

Here, as you are overriding existing members and adding a new one to Iterator[T], the inferred return type of the method it is a structural type of the form Iterator[T] with Object {def next: T; var i : T; def hasNext: Boolean} (as an IDE like IntelliJ can suggest).

Thus you are defining a method whose return type is a structural type that uses an abstract type of this same method. This is what bothers scalac (structural type with same method's abstract type).

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