17
votes

Consider the code below:

trait A {
  def work = { "x" }
}

trait B {
  def work = { 1 }
}

class C extends A with B {
  override def work = super[A].work
}

Class C won't compile in scala 2.10, because of "overriding method work in trait A of type => String; method work has incompatible type".

How to choose one specific method?

4

4 Answers

17
votes

I'm afraid there is no way to do that. The super[A].work way works only if A and B have the same return types.

Consider this:

class D extends B

....

val test: List[B] = List(new C(), new D())
test.map(b => b.work) //oops - C returns a String, D returns an Int
6
votes

Scala simply prevents you from mixing A and B together if they declare a method with the same name and incompatible signature.

6
votes

You can't do that.

Look at this piece of code:

val c = new C
val a: A = c
val b: B = c

There is no way that both of these lines could work:

val s: String = a.work
val i: Int = b.work

If we allowed such code to compile, one of these assignments would have to throw a ClassCastException or fail in another way. So, it simply isn't possible to resolve such conflict.

I guess you have to workaround this with some form of delegation, maybe something like this:

class C extends A {
  def toB = new B {
    //implement B methods by somehow delegating them to C instance
  }
}
6
votes

You can't do that in Scala.

The way to work this around is to use the traits as collaborators

trait A {
  def work = { "x" }
}

trait B {
  def work = { 1 }
}

class C {
  val a = new A { }
  val b = new B { }

  a.work
  b.work
}