实现 trait 的类的类型参数而忽略 trait 参数

Question

I have a trait with type parameters and it has a list containing instances of another classes implementing this trait. I don't care about type paramaters of those instances, so I try to define a type that would allow any subclass implementing the trait with whatever type parameters.

trait Synchronizable[A,B] {
  type S <: Synchronizable[_,_]

  val slaves: MutableList[S]

  def synchronizeWith(q: S) = {
    if (!slaves.contains(q)) slaves += q
  }

}

But it won't compile due to some particular instance doesn't conform to this type definition.

[error]  found   : TwoBuffers.this.faster.type (with underlying type model.collection.SynchronizableTimeSeriesBuffer[someCaseClass ,Option[Any]])
[error]  required: TwoBuffers.this.slower.S
[error]     slower.synchronizeWith(faster)

Answer 1

Firstly, it's not clear from your question how you are using this trait. I think you might want something like this:

trait Synchronizable[A,B] {
  type S = Synchronizable[_, _] // Note =, not <:

  val slaves = MutableList.empty[S] // Create list in trait.

  def synchronizeWith(q: S) = {
    if (!slaves.contains(q)) slaves += q
  }
}

Alternatively, if you really need to use the type S <: Synchronizable[_, _] declaration, then you will need to override the type statement in a sub-class. For example:

trait Synchronizable[A,B] {
  type S <: Synchronizable[_, _]

  val slaves = MutableList.empty[S] // Create list in trait.

  def synchronizeWith(q: S) = {
    if (!slaves.contains(q)) slaves += q
  }
}

class SomeClass[A, B]
extends Synchronizable[A, B] {
  override type S = SomeClass[_, _]
}