Curious whether it would make sense or not to add ...

# arrow

[Scott Christopher]

Curious whether it would make sense or not to add type arguments of the typeclasses to their interface, e.g.

interface Functor<F, A> {
  fun <B> Kind<F, A>.map(f: (A) -> B): Kind<F, B>
}

The reason I ask is that I went to update the Profunctor interface to extend Functor and provide the default implementation for map using rmap, though the only way I could make the types line up was to update Functor<F> to Functor<Kind<F, A>, B> like so:

interface Profunctor<F, A, B> : Functor<Kind<F, A>, B> {
    fun <C, D> Kind2<F, A, B>.dimap(fl: (C) -> A, fr: (B) -> D): Kind2<F, C, D>
    fun <C> Kind2<F, A, B>.lmap(f: (C) -> A): Kind2<F, C, B> = dimap(f, ::identity)
    fun <C> Kind2<F, A, B>.rmap(f: (B) -> C): Kind2<F, A, C> = dimap(::identity, f)
    override fun <C> Kind<Kind<F, A>, B>.map(f: (B) -> C): Kind<Kind<F, A>, C> = rmap(f)
}

[raulraja]

we tried this encoding and also has its shortcomings

[raulraja]

It will be there around April when type proofs are finished

[Jannis]

Btw not sure you are aware of this or even still need it but Kind2<F, A, B> == Kind<Kind<F, A>, B> so Functor<Kind<F, A>> will leave you with <B, C> Kind<Kind<F, A>, B>.map(f: (B) -> C): Kind<Kind<F, A>, C> which is exactly what you need,

[Scott Christopher]

That was what lead me down the path to adding the extra type arguments to the typeclasses, because Profunctor can't extend Functor<Kind<F, A>> without knowing something about A

[Jannis]

Right, forgot 👍
So the only way is to have a method return an instance...