Monoid derivation applies requires Monoid for type parameter even when unnecessary

[mrdziuban]

At least on scala 3 (haven't tested on scala 2) a derived Monoid[F[A]] requires a Monoid[A] even if A is only referenced within types that have a MonoidK instance, i.e. that can form a Monoid regardless of their type parameter.

Could it instead require any number of Monoid[B]s where B is the full type of the each field that references A? For example:

case class Test[A](l: List[A], v: Vector[A]) derives Monoid

// generated before
given derived$Monoid[A](using m: Monoid[A]): Monoid[Test[A]]
// generated after
given derived$Monoid[A](using m1: Monoid[List[A]], m2: Monoid[Vector[A]]): Monoid[Test[A]]

Full example:

https://scastie.scala-lang.org/mrdziuban/wy1ITtp9SCCyUe7uUgfisA/4

import cats.Monoid
import cats.derived.*

case class Test[A](l: List[A]) derives Monoid

Monoid[Test[Boolean]]
/*
No given instance of type cats.kernel.Monoid[Playground.Test[Boolean]] was found for parameter ev of method apply in object Monoid.
I found:

    Playground.Test.derived$Monoid[Boolean](
      /* missing */summon[cats.kernel.Monoid[Boolean]]
    )

But no implicit values were found that match type cats.kernel.Monoid[Boolean].

One of the following imports might fix the problem:

  import cats.derived.auto.commutativeMonoid.given_CommutativeMonoid_A
  import cats.derived.auto.monoid.given_Monoid_A
*/

[joroKr21]

Yeah I think that comes from Scala 3 and we can't control it 😞

[joroKr21]

The workaround is to add the instance to the companion object explicitly:

object Test:
  given [A] Monoid[Test[A]] = semiauto.monoid

[mrdziuban]

Oooh interesting and good to know! Feel free to close this then since it's not a kittens issue

[joroKr21]

Maybe we should document this in the README so I will keep it open.