Chaining \/ with flatMap or >>=

[Karsten Jeschkies]

Hi,

I wanted to get a little acquainted with Scalaz and thought that disjunctions would be a good start. I found that chaining them with flatMap and >>= give to different return types:

import scalaz._, Scalaz._
type ErrorMessage = String

def parse(value: String): ErrorMessage \/ Double =
    value.parseDouble match {
        case Success(x: Double) => \/.right[ErrorMessage, Double](x)
        case _ => \/.left(s"Could not parse '$value'")
    }

def divide(d: Double): ErrorMessage \/ Double =
    if(d == 0.0) \/.left("Cannot divide by 0.0.")
    else \/.right(100 / d)

scala> parse("2.0").flatMap(divide)
res0: scalaz.\/[ErrorMessage,Double] = \/-(50.0)


scala> parse("2.0") >>= divide
res1: scalaz.Unapply[scalaz.Bind,scalaz.\/[ErrorMessage,Double]]{type M[X] = scalaz.\/[ErrorMessage,X]; type A = Double}#M[Double] = \/-(50.0)

I think that >>= is the Monad bind and flatMap is defined in Either.scala as a method of \/. While both have the same behaviour, should both also yield the same type? What am I missing? Also, what is scalaz.Unapply?

Thanks a ton,

Karsten

[Daniel Spiewak]

It's a bit of type-level witchcraft.  They are the same type, just disguised.  You can satisfy yourself of this fact by using the following incantation:

def testEq[A, B](a: A, b: B)(implicit ev: A =:= B): Unit = ()

testEq(parse("2.0").flatMap(divide), parse("2.0") >>= divide)

The above should compile, which is in fact proof that the types are equal.

Daniel

[Alois Cochard]

Hi,

About Unapply, that should help:
http://typelevel.org/blog/2013/09/11/using-scalaz-Unapply.html
http://eed3si9n.com/learning-scalaz/Unapply.html

Cheers

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[Karsten Jeschkies]

Hi,

thanks a lot. That helped somewhat. It still seems that the different methods take different codes paths, though.