scalatypestype-inferenceself-type

How do I make sure a function receives the same parameter type as the current object?


abstract class A {
  protected[this] type ThisType = this.type
  protected[this] type OtherType = this.type 

  def assign(other: OtherType): ThisType
}

class B extends A {
  def assign(other: OtherType): ThisType = ???
}

class C extends A {
  def assign(other: OtherType): ThisType = ???
}

class D extends C {
  def assign(other: OtherType): ThisType = ???
}

How do I make sure the other type received in assign of and object of type B is also B. e.g. how can I write something effectively like:

def f1(p1: A, p2: A) = p1.assign(p2)
def f2[T <: A](p1: T, p2: T) = p1.assign(p2)

I am getting following errors:

Error at p1 Error at p2

NB: Actually ThisType and OtherType should be the same but I seperated them so I can try out different options.


Solution

  • One way of achieving what you want can be done using Type Projections:

    def main(args: Array[String]): T = {
      f1(new C, new C)
    }
    
    abstract class A {
      type ThisType <: A
      def assign(other: ThisType): ThisType
    }
    
    class C extends A {
      override type ThisType = C
      override def assign(other: C): C = ???
    }
    
    class D extends C {
      override type ThisType = D
      override def assign(other: D): D = ???
    }
    
    def f1[T <: A](p1: T#ThisType, p2: T#ThisType) = p1.assign(p2)
    

    Another way can be using F-bound polymorphism:

    abstract class A[T <: A[T]] {
      def assign(other: T): T
    }
    
    class C extends A[C] {
      override def assign(other: C): T = ???
    }
    
    def f1[T <: A[T]](p1: T, p2: T) = p1.assign(p2)