Maulik Soneji - 7 months ago 59

Scala Question

I had a query about what is the difference between parameterising the class vs parameterising the function.

I have provided implementation of a Functor as follows:

`trait Functor[F[_],A,B] {`

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

}

and the other where the function is parameterised as follows:

`trait Functor[F[_]] {`

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

}

Where are the cases where we should use one over another?

Another follow up question:

Why do we pass the argument to functor as F[_] and not as F[A] or F[B]. What cases arise when we use either F[A] or F[B]?

Answer

**Always** prefer the second one. With the first one you can implement instances as nonsensical as:

```
trait WrongFunctor[F[_],A,B] {
def map(fa: F[A])(f: A => B) : F[B]
}
case class notEvenRemotelyAFunctor[A]() extends WrongFunctor[List,A,Int] {
def map(fa: List[A])(f: A => Int) : List[Int] =
if(f(fa.head) < 4) List(3) else List(4)
}
type Id[X] = X
case object ILikeThree extends WrongFunctor[Id, Int, Int] {
def map(fa: Int)(f: Int => Int): Int = if(fa == 3) 3 else f(fa)
}
```

Even if you do things right, you'd need for a fixed functor implementation one object per different types at which you want to use `fmap`

. But the important point is that the second one at least makes it harder to write that kind of wrong "functors"; less non-functors will slip by:

```
trait Functor[F[_]] {
def map[A,B](fa: F[A])(f: A => B) : F[B]
}
case object ILikeThreeAgain extends Functor[Id] {
def map[A,B](fa: A)(f: A => B) : B =
??? // how do I write the above here?
}
```

The keywords here are *parametricity* and *parametric polymorphism*. The intuition is that if something is defined generically, you can derive properties it will satisfy just from the types involved. See for example Bartosz Milewski blog - Parametricity: Money for Nothing and Theorems for Free for a good explanation, or the canonical Theorems for free paper.

Another follow up question: Why do we pass the argument to functor as F[_] and not as F[A] or F[B]. What cases arise when we use either F[A] or F[B]?

Because that's part of what a functor is; it is a "constructor":

- for each input type
`A`

it gives you as output another type`F[A]`

- and for each function
`f: A => B`

, another function`fmap(f): F[A] => F[B]`

satisfying`fmap(id[A]) == id[F[A]]`

and`fmap(f andThen g) == fmap(f) andThen fmap(g)`

So for 1. you need a way of representing functions *on* types; and that's what `F[_]`

is.

Note that having a `map`

method like in your signature is in this case equivalent to `fmap`

:

```
trait Functor[F[_]] {
def map[A,B](fa: F[A])(f: A => B) : F[B]
def fmap[A,B](f: A => B): F[A] => F[B] =
{ fa => map(fa)(f) }
def mapAgain[A,B](fa: F[A])(f: A => B) : F[B] =
fmap(f)(fa)
}
```

Now for how this links with real category theory:

Instances of your `Functor[F[_]]`

trait above are meant to represent **Scala**-enriched functors

F: **Scala** → **Scala**

Let's unpack this.

There's a (usually implicitly defined) category **Scala** with objects types, and morphisms functions f: A ⇒ B. This category is cartesian closed, where the internal hom is the *type* A ⇒ B, and the product (A,B). We can work then with **Scala**-enriched categories and functors. What's a **Scala**-enriched category? basically one that you can *define* using Scala the language: you have

- a set of objects (which you'd need to represent as types)
- for each A,B a type
`C[A,B]`

with identities`id[X]: C[X,Y]`

and composition`andThen[X,Y,Z]: (C[X,Y], C[Y,Z]) => C[X,Z]`

satisfying the category axioms

Enriched functors F: C → D are then

- a function from the objects of C to those of D,
`A -> F[A]`

- for each pair of objects A,B: C a morphism in
**Scala**i.e. a function`fmap: C[A,B] => C[F[A], F[B]]`

satisfying the functor laws`fmap(id[A]) == id[F[A]]`

and`fmap(f andThen g) == fmap(f) andThen fmap(g)`

**Scala** is naturally enriched in itself, with `Scala[X,Y] = X => Y`

, and enriched functors F: **Scala** → **Scala** are what instances of your `Functor[F[_]]`

trait are meant to represent.

Of course this needs all sort of qualifications about how Scala breaks this and that, morphism equality etc. But the moral of the story is: your base language **L** (like Scala in this case) is likely trying to be a cartesian-closed (or at least symmetric monoidal-closed) category, and functors definable through it correspond to **L**-enriched functors.

Source (Stackoverflow)