How to specifiy shared super type as parameter in Scala?

Question

I need some help on this scenario:

A trait defines an shared interface with some methods. One of the methods should look like this

 def fromConfig( config : Config): 

Each implementation should override the method with something like:

  override def fromConfig(config: SubConfigA)

However, doing so gives "Method .. overrides nothing" error
even though the config stuff is defined as

  abstract class Config
  trait SubConfigA extends Config
  trait SubConfigB extends Config 
  ....

I remember, in Java you actually can define an interface as parameter, and use sub-interfaces in subclasses for refinement.

Apparently the same does not work in Scala.

The code above is just a draft but the idea is passing specific configurations matching to a specific implementation of the shared interface.

Any idea or advice would be most appreciated.

Thank you

Answer 1

Is this what you want?

abstract class Config
trait ConfigA extends Config
trait ConfigB extends Config

trait FromConfigLoader[C <: Config] {
  def fromConfig(config: C)
}

class TestA extends FromConfigLoader[ConfigA] {
 def fromConfig(config: ConfigA) = ???
}

class TestB extends FromConfigLoader[ConfigB] {
 def fromConfig(config: ConfigB) = ???
}