# Typeclases in Scala & Haskell

# Typeclases in Scala & Haskell

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Join For FreeTypeclass is a concept in Scala and Haskell (and in some other strongly-typed languages). The best way to think about a typeclass is that it is an interface, which defines functions. Great. Typically, everyone really understands this. The trouble comes when we have to apply the concept to practical code, especially if we’ve grown up with Scala and we now want to try out Haskel.

As I said above, typeclass is essentially an interface that defines some behaviour. Consider a `Monoid`

typeclass for some type `a`

. In Haskell, one writes:

class Monoid a where mempty :: a mappend :: a -> a -> a

This is the equivalent of a `trait`

in Scala parametrized with type variable:

trait Monoid[A] { def mempty: A def mappend(a1: A, a2: A): A }

If you look at these typeclass definitions, you should see that these are just the interfaces. They contain no behaviour. Let’s now find out how to define instances of typeclasses; a typeclass instance is an implementation of the interface (say, the `Monoid`

above) for a specific type. Consider:

instance Monid Int where mempty = 0 mappend a b = a + b

Of course, being Haskell, we can eliminate the values in `mappend`

and write just `mappend = (+)`

. Now, in Scala, we can have:

class IntMonoid extends Monoid[Int] { def mempty = 0 def mappend(a: Int, b: Int) = a + b }

If we now define a polymorphic function (with respect to its parameters), we can require that there be an instance of typeclass for the type.

sigma :: (Monoid a) => Int -> Int -> (Int -> Int) -> (Int -> a) -> a sigma a b inc comp = if a > b then mempty else mappend (comp a) (sigma (inc a) b inc comp)

Before I dig into the details, let me show equivalent code in Scala:

def sigma[A](a: Int, b: Int, inc: Int => Int, comp: Int => a) (implicit m: Monoid[A]): A = if (a > b) m.mempty else m.append(comp(a), sigma(inc(a), b, inc, comp))

So, in Scala, we have to be a bit more explicit; we need to tell the compiler which monoid instance (instance here is the OO instance, a value that implements `Monoid[A]`

). However, the compiler can search the *implicit scope* and find an instance of `Monoid[Int]`

; of course reporting errors if it cannot find exactly one instance.

Now, `class IntMonoid extends Monoid[Int]`

is not an instance. It is a class definition, and therefore the compiler will complain that there is no implicit instance of `Monoid[Int]`

. In Scala, we have to provide an implicit value. We can do so by writing:

implicit val intMonoid = new IntMonoid

Or, in this particular case, we can actually make the `Monoid[Int]`

a singleton!

implicit object IntMonoid extends Monoid[Int] { def mempty = 0 def mappend(a: Int, b: Int) = a + b }

If this `object IntMonoid`

is in the current scope, the Scala compiler will be able to supply it as the value of the `implicit m: Monoid[A]`

parameter; and our function compiles & runs as expected.

### Haskell for great good

Compare the code in Haskell:

sigma :: (Monoid a) => Int -> Int -> (Int -> Int) -> (Int -> a) -> a sigma a b inc comp = if a > b then mempty else mappend ...

And the Scala code

// option 1: explicit implicits def sigma[A](a: Int, b: Int, inc: Int => Int, comp: Int => A) (implicit m: Monoid[A]): A = if (a > b) m.mempty else m.append(...) // option 2.1: type bounds def sigma[A : Monoid](a: Int, b: Int, inc: Int => Int, comp: Int => A): A = if (a > b) implicitly[Monoid[A]].mempty else implicitly[Monoid[A]].append(...) // option 2.2: type bounds and then consume implicit def sigma[A : Monoid](a: Int, b: Int, inc: Int => Int, comp: Int => A): A = { val m = implicitly[Monoid[A]] if (a > b) m.mempty else m.mappend(...) }

Because Scala is OO, there are more ways in which you provide the typeclass instances; even though ultimately, you have to have an implicit value available in the implicit scope. You can, however write

// implicit singleton value implicit object IntMonoid extends Monoid[Int] { ... } // implicit anonymous implementation implicit val intMonoid = new Monoid[Int] { ... } // implicit function that returns a new Monoid[Int] every time implicit def intMonoid = new Monid[Int] { ... } // implicit named implementation class IntMonoid extends Monoid[Int] { ... } implicit val intMonoid = new IntMonoid

One might now think that in Haskell, the typeclass instances are equivalent of `implicit object`

construct in Scala. Well, there are some terms & conditions!

### Typeclass “inheritance”

We can have a kind of inheritance in our typeclasses. Let’s add typeclass for `Group`

.

class Group a where mempty :: a mappend :: a -> a -> a inverse :: a -> a

This feels like too much typing. We would really like to “inherit” all functions from `Monoid a`

and only provide the `inverse`

. Our “inheritance” concept say that if I have an instance of `Group`

for `a`

, then I also have an instance of `Monoid`

for the same `a`

. It is not the traditional OO subclassing situation.

class (Monoid a) => Group a where inverse :: a -> a

Here, we have class `Group`

for some type `a`

, that “inherits” functions from `Monoid`

for the same type `a`

. To create instance of `Group Int`

, we can now only provide the one remaining function.

instance Group Int where inverse = ((-1) *)

And we can now happily require these instances:

example :: (Group a) => a -> a example a = inverse (mappend a a)

We can do the same thing in Scala, of course; though the syntax is a little more hairy, and sometimes, we have to deal with unpleasant side-effects. Let’s define `Group[A]`

, which extends `Monoid[A]`

.

trait Group[A] extends Monoid[A] { def inverse(a: A): A }

Now, how do we go about creating instances of `Group[A]`

if we already have `Monoid[A]`

? We could, of course write the whole thing:

implicit object IntGroup extends Group[Int] { def mempty = 0 def mappend(a: Int, b: Int) = a + b def inverse(a: Int) = -a }

Or (inefficiently!) we can write code that’s similar to the Haskell code:

implicit def intGroup(implicit m: Monoid[Int]): Group[Int] = new Group[Int] { def mempty = m.mempty def mappend = m.mappend def invserse(a: Int) = -a }

Here you can see that the compiler will create an instance of `Group[Int]`

, as long as it already has an instance of `Monoid[Int]`

. This is where Haskell’s typeclasses really shine; Scala’s approach is not as efficient or elegant.

### The epiphany

On a recent training ride, a friend told me, “I get typeclasses in Scala, but how can I call functions on the typeclass instances in Haskell?” Let me go back to the `sigma`

function. In Scala, we write

def sigma[A](a: Int, b: Int, inc: Int => Int, comp: Int => a) (implicit m: Monoid[A]): A = if (a > b) m.mempty else m.append(comp(a), sigma(inc(a), b, inc, comp))

In Haskell, we write:

sigma :: (Monoid a) => Int -> Int -> (Int -> Int) -> (Int -> a) -> a sigma a b inc comp = if a > b then mempty else mappend (comp a) (sigma (inc a) b inc comp)

*The difference is that in Haskell, the compiler makes all functions from the typeclass available in the body of the sigma function; and it *knows

*which one to call based on the types.*

Published at DZone with permission of Jan Machacek , DZone MVB. See the original article here.

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