In the following code snippet, I define a Function1 Functor:
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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. |
W dniu środa, 19 kwietnia 2017 09:16:59 UTC+2 użytkownik Joe San napisał:
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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.
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.
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. |
In reply to this post by Joe San
Op woensdag 19 april 2017 09:16:59 UTC+2 schreef Joe San:
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.
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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:
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In reply to this post by Joe San
On 4/19/17 3:16 AM, Joe San wrote:
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.
Not any type, but lots of them, sure. Including the one below.
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. |
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