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How to use >=> in Scala?

I am trying to use >=> (Kleisli arrow) in Scala. As I understand, it composes functions returning monads. Now I am trying it as follows:

scala> val f = {i:Int => Some(i + 1)}
f: Int => Some[Int] = <function1>

scala> val g = {i:Int => Some(i.toString)}
g: Int => Some[String] = <function1>

scala> val h = f >=> g
<console>:15: error: value >=> is not a member of Int => Some[Int]
       val h = f >=> g
                 ^

Why does not it compile ? How to compose f and g with >=> ?

like image 602
Michael Avatar asked Feb 09 '14 14:02

Michael


1 Answers

There are two problems here. The first is that the inferred types of your functions are too specific. Option is a monad, but Some is not. In languages like Haskell the equivalent of Some isn't even a type—it's just a constructor—but because of the way algebraic data types are encoded in Scala you have to watch out for this issue. There are two easy fixes—either provide the more general type explicitly:

scala> val f: Int => Option[Int] = i => Some(i + 1)
f: Int => Option[Int] = <function1>

scala> val g: Int => Option[String] = i => Some(i.toString)
g: Int => Option[String] = <function1>

Or use Scalaz's handy some, which returns an appropriately typed Some:

scala> val f = (i: Int) => some(i + 1)
f: Int => Option[Int] = <function1>

scala> val g = (i: Int) => some(i.toString)
g: Int => Option[String] = <function1>

The second problem is that >=> isn't provided for plain old monadic functions in Scalaz—you need to use the Kleisli wrapper:

scala> val h = Kleisli(f) >=> Kleisli(g)
h: scalaz.Kleisli[Option,Int,String] = Kleisli(<function1>)

This does exactly what you want—just use h.run to unwrap.

like image 188
Travis Brown Avatar answered Oct 05 '22 03:10

Travis Brown