# Function Functors Definition

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## Function Functors Definition

 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? -- You received this message because you are subscribed to the Google Groups "scala-user" group. To unsubscribe from this group and stop receiving emails from it, send an email to [hidden email]. For more options, visit https://groups.google.com/d/optout.
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## Re: Function Functors Definition

 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   -- You received this message because you are subscribed to the Google Groups "scala-user" group. To unsubscribe from this group and stop receiving emails from it, send an email to [hidden email]. For more options, visit https://groups.google.com/d/optout.
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## Re: Function Functors Definition

 In reply to this post by Joe San 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. 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? -- You received this message because you are subscribed to the Google Groups "scala-user" group. To unsubscribe from this group and stop receiving emails from it, send an email to [hidden email]. For more options, visit https://groups.google.com/d/optout.
 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: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. 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? -- You received this message because you are subscribed to the Google Groups "scala-user" group. To unsubscribe from this group and stop receiving emails from it, send an email to [hidden email]. For more options, visit https://groups.google.com/d/optout.
 In reply to this post by Joe San 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 -- You received this message because you are subscribed to the Google Groups "scala-user" group. To unsubscribe from this group and stop receiving emails from it, send an email to [hidden email]. For more options, visit https://groups.google.com/d/optout.