问题
Let's say I have two optional Ints (both can be Some or None):
val one : Option[Int] = Some(1)
val two : Option[Int] = Some(2)
My question is the following: Are there any intelligent way to sum them op using Scalas brilliant collection-methods? I realize that I could merge them into a collection, flatten it and use reduceLeftOption like so:
(one :: two :: Nil).flatten.reduceLeftOption(_ + _) // Some(3)
But, the solution above means creating a new collection, and living in a rich and developed world that takes time from all the other first world activities I might immerse myself into. And in a world where programming gets more and more luxurious for programmers like us, there must be one or more luxurious first world answer(s) to this, right?
Edit: So to spell things out, here are some examples:
If one = Some(1) and two = Some(2) we should have Some(3)
If one = Some(1) and two = None we should have Some(1)
If one = None and two = Some(2) we should have Some(2)
If both one and two are None we should have None, since neither one or two can be summed correctly.
Hope that clarified things :-)
回答1:
obligatory scalaz answer is to use the scalaz Option monoid:
scala> one |+| two
res0: Option[Int] = Some(3)
It will do what you want with respect to None:
scala> two |+| None
res1: Option[Int] = Some(2)
scala> none[Int] |+| none[Int]
res2: Option[Int] = None
That none method is a method from scalaz which helps with type inference because instead of returning None <: Option[Nothing] it returns a Option[Int], there is a similar method from Some which returns an Option[A] for any given A instead of a Some[A]:
scala> 1.some |+| 2.some
res3: Option[Int] = Some(3)
回答2:
for (x <-one; y <- two) yield x+y
Or the less readable but strictly equivalent:
one.flatMap{x=>two.map(x+_)}
UPDATE: As your latest edit made quite clear, you only want a None as the result when both the input options are None. In this case I don't think you'll get anything better in terms of simplicity than what you already use. I could shorten it a bit but overall this is jsut the same:
(one ++ two).reduceOption(_ + _)
回答3:
How about:
one.map(_ + two.getOrElse(0)).orElse(two)
回答4:
You could try this:
for( x <- one.orElse(Some(0)); y <- two.orElse(Some(0))) yield x+y
来源：https://stackoverflow.com/questions/16319505/summing-up-two-options