scalapolymorphismtraitsassociated-typesf-bounded-polymorphism

How to combine F-bounded polymorphism with associated types in Scala?


I have a trait called Graphlike for things that work as a graph. Notably, one of the properties I want to have is that the method g.subgraph(Set(1, 2, 3)) would return a subgraph of the same type with just the vertices 1, 2 and 3. Apparently, this means that I want F-bounded polymorphism and that Graphlike looks something like this:

trait Graphlike[A <: Graphlike[A]] {
  type Vertex

  def subgraph(selectedVertices: Set[Vertex]): A
}

I also have a trait representing an automaton with associated types for edges and vertices. I want it to behave like a graph. Simplified, it looks like this:

trait Automaton extends Graphlike[Automaton] {
  type State
  type Vertex = State

  def states: Iterable[State]
  def initialState: State
}

This almost works. However, Scala's type system gets confused when I try to mix the two and do something useful with the result:


class UsesAutomataAsGraphs(val aut: Automaton with Graphlike[Automaton]) {
  aut.subgraph(Set(aut.initialState)).subgraph(Set(aut.initialState))
}

gives an error like:

[info] Compiling 1 Scala source to /Users/albin/Downloads/example/target/scala-2.12/classes ...
[error] /Users/albin/Downloads/example/src/main/scala/example/Example.scala:21:56: type mismatch;
[error]  found   : UsesAutomataAsGraphs.this.aut.State
[error]  required: _1.State where val _1: Automaton
[error]   aut.subgraph(Set(aut.initialState)).subgraph(Set(aut.initialState))

How do I get Scala to understand that the two associated types are the same in all derived objects?


Solution

  • The easiest solution seems to be this. Just make sure that subgraph returns the same type with this.type and you're good to go. There's no need for A - it just adds additional complexity as you try to prove that A is the type of this.

    trait Graphlike {
      type Vertex
    
      def subgraph(selectedVertices: Set[Vertex]): this.type
    }
    
    trait Automaton extends Graphlike {
      type State
      type Vertex = State
    
      def states: Iterable[State]
      def initialState: State
    }
    
    class UsesAutomataAsGraphs(val aut: Automaton) {
      aut.subgraph(Set(aut.initialState)).subgraph(Set(aut.initialState))
    }
    

    In Scastie: https://scastie.scala-lang.org/zMtde7VISKi18LdPXO6Ytw



    Having State be a type parameter also worked for me. Note that in UsesAutomataAsGraphs, if you use A <: Automaton[_] (a wildcard), it doesn't work, because State could be anything. The compiler wants your assurance that the returned Automaton will have the same State type (because it's unbounded, and other classes extending Automaton could define it differently).

    trait Graphlike[A <: Graphlike[A]] {
      type Vertex
    
      def subgraph(selectedVertices: Set[Vertex]): A
    }
    
    trait Automaton[State] extends Graphlike[Automaton[State]] {
      type Vertex = State
    
      def states: Iterable[State]
      def initialState: State
    }
    
    class UsesAutomataAsGraphs[S](val aut: Automaton[S]) {
      aut.subgraph(Set(aut.initialState)).subgraph(Set(aut.initialState))
    }
    

    Link to Scastie: https://scastie.scala-lang.org/RolPc3ggTxeZ2tUqdXKNEQ


    It also works if you define subgraph as such:

    def subgraph(selectedVertices: Set[_ >: Vertex]): this.type
    

    Because it's contravariant, even if Vertex is different in different classes and/or traits, it'll work.

    Link to Scastie: https://scastie.scala-lang.org/fz509HEpTBGoJGaJxLziBQ