CODEWITHSUNDEEP
scalafunctional-programmingscalaz
Xiaohe Dong
Xiaohe Dong

Reputation: 5023

How to flip Seq[Free[Functor, A]] to Free[Functor, Seq[A]]

I am learning scalaz Free monad. I wonder how to can I get Seq[Free[Functor, A]] to Free[Functor, Seq[A]] similar like Future.sequence

Here is the code 

  type Functor[A] = Coyoneda[Command, A]
  type Script[A]= Free[Functor, A]

  case instruction(...)

  def sequence(): Seq[Script[Instruction]] = {...}

I wonder that is that possible that I can flip the result to Script[Seq[Instruction]] If it can be like Future does, Is that the reason Free is a monad ?

Many thanks in advance

Upvotes: 2

Views: 322

Answers (2)

Tomas Mikula
Tomas Mikula

Reputation: 6537

There is the sequence operation for F[G[A]] (e.g. Seq[Script[Instruction]]) whenever there are Traverse[F] and Applicative[G] instances. You do have the Applicative[Script] instance (Script is a Monad, so it is also Applicative), but scalaz does not provide a Traverse instance for Seq. (Immutable Seq is in principle traversable, but the authors chose not to provide Traverse[Seq] instance, presumably because Seq is abstract and hence scalaz cannot guarantee to give you back the same Seq implementation.)

I will therefore answer for List instead of Seq.

import scalaz.std.list._
import scalaz.syntax.traverse._

val scripts: List[Script[A]] = ???
val script: Script[List[A]] = scripts.sequence[Script, A]

Upvotes: 1

Raúl Raja Martínez
Raúl Raja Martínez

Reputation: 684

Free.freeMonad#sequence will help you flip the type constructors in both Cats and Scalaz. Alternative require the monad instance via implicit evidence and use the compiler injected implicit instance instead.

Upvotes: 0

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