Scala: How to define an enum with extra attributes? - scala

Question

I have a use-case where I need to define a new enum type LongShort but I need it in a way to also carry the sign so it can be directly used in mathematical expressions e.g.
object LongShortType extends Enumeration {
type Type = Value
val Long = Value(+1)
val Short = Value(-1)
}
I'd then like to use it like this:
val longShort = LongShortType.Short
val numberOfContracts: Int = 10
val vanillaOptionNotional: Double = longShort*numberOfContracts
but this leads to compiler error cannot resolve symbol * ... is there a way to extract the value of the enum? Or am I not understanding how enum types work?

Answer 1

The type of LongShortType.Short isn't Int, it's Value. You can either extract the underlying id of the value:
val longShort = LongShortType.Short.id
Which is a little ugly. Or you could not use an enum type at all:
object LongShortType {
val Long = 1
val Short = -1
}
And then your equation would work as is.

Answer 2

OK I worked out a solution to accomplish what I wanted without any compromisse and by that I mean that this solution has all the advantages of using Scala enum e.g. the withName and still allows me to define extra features on it:
object LongShortType extends Enumeration {
type Type = LongShortVal
val Long = Value("Long", +1)
val Short = Value("Short", -1)
case class LongShortVal(name: String, sign: Int) extends Val(nextId, name)
protected final def Value(name: String, sign: Int) = new LongShortVal(name, sign)
}
and now can do:
val longShort = LongShortType.Short
val numberOfContracts: Int = 10
val vanillaOptionNotional: Double = longShort.sign*numberOfContracts
and can also do:
val longShort = LongShort.withName("Long") // returns LongShort.Long