How to go from an functor over an effect to an effect over the functor?

Question

I have two different types of users; let's call them BasicUser and EnrichedUser.

I get a value of the Foo user type inside a container C that has a cats.Functor instance:

val basicUserF: C[BasicUser] = getBasicUser(…)

Now I'd like to map over this value (using the functor instance) to turn the contained BasicUser into an EnrichedUser user, staying in the container:

import cats.syntax.all._

val enrichedUserF: C[EnrichedUser] = basicUserF.map(user => enrichUser(user))

However, enrichUser doesn't return EnrichedUser, but rather OptionT[Future, EnrichedUser], so I end up with the following type:

val enrichedUserThing: C[OptionT[Future, EnrichedUser]] = …

How can I get from this type to OptionT[Future, C[EnrichedUser]] (which I'll ultimately turn into a JSON-serialized Akka HTTP response)?

Dies this help https://stackoverflow.com/questions/20874186/scala-listfuture-to-futurelist-disregarding-failed-futures ? — n. m. could be an AI, Jun 01 '17 at 13:19
@n.m. I'm not sure. I can't see how Future.sequence could help me… — , Jun 01 '17 at 13:22
If you add a `Traverse` constraint to your `C`, you could use `sequence` which would do what you want. — Luka Jacobowitz, Jun 01 '17 at 13:25

Jasper-M · Accepted Answer · 2017-06-01T13:41:46.487

1

If you have a Traverse[C] instance you can use sequence.

scala> implicit def CTraverse: Traverse[C] = ???
CTraverse: cats.Traverse[C]

scala> def sequenced = enrichedUserThing.sequence[OptionT[Future,?], EnrichedUser]
sequenced: cats.data.OptionT[scala.concurrent.Future,C[EnrichedUser]]

Extra: If you have -Ypartial-unification enabled you don't need the explicit type arguments.

edited Jun 01 '17 at 13:41

answered Jun 01 '17 at 13:25

Jasper-M

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Okay, so I need to bite the bullet and add a Traverse instance to C :| Thanks :) – Jun 01 '17 at 13:41

How to go from an functor over an effect to an effect over the functor?

1 Answers1