CODEWITHSUNDEEP
scala
Chris Bogart
Chris Bogart

Reputation: 472

Can I declare a decorator class that extends its own type parameter?

I'd like to do something like this:

case class D[X <: A](arg1 : X, arg2: Int) extends X {
}

D is kind of a decorator class for arg1, and I'd like to apply it to several different kinds of things that are subclasses of A.

However I get this error:

scala> case class D[X <: A](arg1 : X, arg2: Int) extends X { override val name = "D"; } :6: error: class type required but X found

If not, is there a more scalaish way to do this kind of thing?

Upvotes: 1

Views: 430

Answers (1)

axel22
axel22

Reputation: 32345

The class that you extend has to be known at compile time and a type parameter is generally not. Therefore, it's not possible to do this.

However, if you're trying to extend X to benefit from the implementations of methods defined in an interface trait A, then you can mix-in X when instantiating the class.

new D with X

If you'd like to preserve the 'case class' features of D, then using D as a proxy which forwards calls to methods defined in A to the parameter arg1 of type X is one solution.

trait A {
  def foo
}
case class D[X <: A](arg1: X) extends A {
  def forw = arg1
  def foo = forw.foo
}

Upvotes: 1

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