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