Function Functors Definition

[Joe San]

In the following code snippet, I define a Function1 Functor:

implicit def Function1Functor[R]: Functor[({type l[a]=(R) => a})#l] = new Functor[({type l[a]=(R) => a})#l] {
  def fmap[A, B](r: R => A, f: A => B) = r andThen f
}

1. But why is that signature ({type l[a]=(R) => a})#l like that? how to interpret that?

2. I could now randomly pick any type and qualify that as a Functor. For example.,

 

     case class SomeRandomType[A] (param: A) // this does not know about the fact that it is going to be a Functor soon

     val someRandomTypeFunctor = new Functor[SomeRandomType] {

       def map[A, B](fa: SomeRandomType[A])(f: A => B): SomeRandomType[B] = 

          SomeRandomType(f(fa.param))

     }

 

So this to me is a Typeclass definition. Is that correct?

[Rafał Krzewski]

W dniu środa, 19 kwietnia 2017 09:16:59 UTC+2 użytkownik Joe San napisał:

In the following code snippet, I define a Function1 Functor:

implicit def Function1Functor[R]: Functor[({type l[a]=(R) => a})#l] = new Functor[({type l[a]=(R) => a})#l] {
  def fmap[A, B](r: R => A, f: A => B) = r andThen f
}

1. But why is that signature ({type l[a]=(R) => a})#l like that? how to interpret that?

This is called a type lambda. You can think about it as an anonymous function in the domain of types. It is typically used in Scala to partially apply a type constructor, ie. convert a type constructor that takes more parameters (Function1[-A,+R] in your case) to a type constructor that takes less parameters (Function1Functor[A] respectively). Google it up, there are some really good articles about it out there.

 

2. I could now randomly pick any type and qualify that as a Functor. For example.,

 

Well, for any type constructor F[_] an instance of Functor[F] can be summoned, using Yoneda lemma. But this is advanced category-theoretical wizardry, on the far boundary of my expertise. There are also materials and talks out there.

Google up "free monads", because deriving Functors for arbitrary F[_] is an essential step in this technique.

 

     case class SomeRandomType[A] (param: A) // this does not know about the fact that it is going to be a Functor soon

     val someRandomTypeFunctor = new Functor[SomeRandomType] {

       def map[A, B](fa: SomeRandomType[A])(f: A => B): SomeRandomType[B] = 

          SomeRandomType(f(fa.param))

     }

 

So this to me is a Typeclass definition. Is that correct?

It appears to be, yes. Unfortunately I don't know if it's consistent, or the correct alignment of types there is a mere coincidence ;)

cheers,

Rafał 

 

[Jasper-M]

Op woensdag 19 april 2017 09:16:59 UTC+2 schreef Joe San:

In the following code snippet, I define a Function1 Functor:

implicit def Function1Functor[R]: Functor[({type l[a]=(R) => a})#l] = new Functor[({type l[a]=(R) => a})#l] {
  def fmap[A, B](r: R => A, f: A => B) = r andThen f
}

1. But why is that signature ({type l[a]=(R) => a})#l like that? how to interpret that?

Functor is defined like this (more or less): trait Functor[F[_]].  This means that it accepts a type constructor of arity 1.

Function1 is defined like this: trait Function1[-A,+B]. So Function1 is a type constructor of arity 2.

If you write a Functor for List (arity 1) you can just say Functor[List] and you're done. But Functor[Function1] is impossible.

So to create a Functor for a Function1 you have to supply 1 type argument to Function1 and leave 1 parameter open, or in other words partially apply the Function1 type constructor.

Scala doesn't have special syntax for partially applying type constructors, so you have to do it with ({ type Lambda[x] = Function1[SomeType,x] })#Lambda.

How this works is that you create a structural type that contains a type alias called Lambda which is a type constructor of arity 1. Then you use the # operator on that structural type to extract the Lambda type constructor; this is called type projection.

[Jasper-M]

There is a compiler plugin (https://github.com/non/kind-projector) which adds special syntax for this purpose.

It enables you to write Functor[R => ?].

Op woensdag 19 april 2017 11:08:17 UTC+2 schreef Jasper-M:

[Stephen Compall]

On 4/19/17 3:16 AM, Joe San wrote:

1. But why is that signature ({type l[a]=(R) => a})#l like that? how to interpret that?

Given

type l[a]=(R) => a

then

l

Or, alternatively, you can think of it as meaning

l

where

type l[a]=(R) => a

Jasper has described the mechanism by which this is implemented.

2. I could now randomly pick any type and qualify that as a Functor.

Not any type, but lots of them, sure.  Including the one below.

For example.,

 

     case class SomeRandomType[A] (param: A) // this does not know about the fact that it is going to be a Functor soon

     val someRandomTypeFunctor = new Functor[SomeRandomType] {

       def map[A, B](fa: SomeRandomType[A])(f: A => B): SomeRandomType[B] = 

          SomeRandomType(f(fa.param))

     }

 

So this to me is a Typeclass definition. Is that correct?

Yep.


--
Stephen Compall