And now, we’re going to take a look at the Functor typeclass, which is basically for things that can be mapped over.

Like the book let’s look how it's implemented:

 * Functor.
 * The name is short for "covariant functor".
 * Must obey the laws defined in cats.laws.FunctorLaws.
@typeclass trait Functor[F[_]] extends functor.Invariant[F] { self =>
  def map[A, B](fa: F[A])(f: A => B): F[B]


Here’s how we can use this:

scala> import cats._,, cats.implicits._
import cats._
import cats.implicits._

scala> Functor[List].map(List(1, 2, 3)) { _ + 1 }
res0: List[Int] = List(2, 3, 4)

Let’s call the above usage the function syntax.

We now know that @typeclass annotation will automatically turn a map function into a map operator. The fa part turns into the this of the method, and the second parameter list will now be the parameter list of map operator:

// Supposed generated code
object Functor {
  trait Ops[F[_], A] {
    def typeClassInstance: Functor[F]
    def self: F[A]
    def map[B](f: A => B): F[B] =

This looks almost like the map method on Scala collection library, except this map doesn’t do the CanBuildFrom auto conversion.

Either as a functor 

Cats defines a Functor instance for Either[A, B].

scala> (Right(1): Either[String, Int]) map { _ + 1 }
res1: Either[String,Int] = Right(2)

scala> (Left("boom!"): Either[String, Int]) map { _ + 1 }
res2: Either[String,Int] = Left(boom!)

Note that the above demonstration only works because Either[A, B] at the moment does not implement its own map. If I used List(1, 2, 3) it will call List’s implementation of map instead of the Functor[List]’s map. Therefore, even though the operator syntax looks familiar, we should either avoid using it unless you’re sure that standard library doesn’t implement the map or you’re using it from a polymorphic function. One workaround is to opt for the function syntax.

Function as a functor 

Cats also defines a Functor instance for Function1.

scala> val h = ((x: Int) => x + 1) map {_ * 7}
h: Int => Int = <function1>

scala> h(3)
res3: Int = 28

This is interesting. Basically map gives us a way to compose functions, except the order is in reverse from f compose g. Another way of looking at Function1 is that it’s an infinite map from the domain to the range. Now let’s skip the input and output stuff and go to Functors, Applicative Functors and Monoids.

How are functions functors? …

What does the type fmap :: (a -> b) -> (r -> a) -> (r -> b) for this instance tell us? Well, we see that it takes a function from a to b and a function from r to a and returns a function from r to b. Does this remind you of anything? Yes! Function composition!

Oh man, LYAHFGG came to the same conclusion as I did about the function composition. But wait…

ghci> fmap (*3) (+100) 1
ghci> (*3) . (+100) $ 1

In Haskell, the fmap seems to be working in the same order as f compose g. Let’s check in Scala using the same numbers:

scala> (((_: Int) * 3) map {_ + 100}) (1)
res4: Int = 103

Something is not right. Let’s compare the declaration of fmap and Cats’ map function:

fmap :: (a -> b) -> f a -> f b

and here’s Cats:

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

So the order is flipped. Here’s Paolo Giarrusso (@blaisorblade)’s explanation:

That’s a common Haskell-vs-Scala difference.

In Haskell, to help with point-free programming, the “data” argument usually comes last. For instance, I can write map f . map g . map h and get a list transformer, because the argument order is map f list. (Incidentally, map is an restriction of fmap to the List functor).

In Scala instead, the “data” argument is usually the receiver. That’s often also important to help type inference, so defining map as a method on functions would not bring you very far: think the mess Scala type inference would make of (x => x + 1) map List(1, 2, 3).

This seems to be the popular explanation.

Lifting a function 


[We can think of fmap as] a function that takes a function and returns a new function that’s just like the old one, only it takes a functor as a parameter and returns a functor as the result. It takes an a -> b function and returns a function f a -> f b. This is called lifting a function.

ghci> :t fmap (*2)  
fmap (*2) :: (Num a, Functor f) => f a -> f a  
ghci> :t fmap (replicate 3)  
fmap (replicate 3) :: (Functor f) => f a -> f [a]  

If the parameter order has been flipped, are we going to miss out on this lifting goodness? Fortunately, Cats implements derived functions under the Functor typeclass:

@typeclass trait Functor[F[_]] extends functor.Invariant[F] { self =>
  def map[A, B](fa: F[A])(f: A => B): F[B]


  // derived methods

   * Lift a function f to operate on Functors
  def lift[A, B](f: A => B): F[A] => F[B] = map(_)(f)

   * Empty the fa of the values, preserving the structure
  def void[A](fa: F[A]): F[Unit] = map(fa)(_ => ())

   * Tuple the values in fa with the result of applying a function
   * with the value
  def fproduct[A, B](fa: F[A])(f: A => B): F[(A, B)] = map(fa)(a => a -> f(a))

   * Replaces the `A` value in `F[A]` with the supplied value.
  def as[A, B](fa: F[A], b: B): F[B] = map(fa)(_ => b)

As you see, we have lift!

scala> val lifted = Functor[List].lift {(_: Int) * 2}
lifted: List[Int] => List[Int] = <function1>

scala> lifted(List(1, 2, 3))
res5: List[Int] = List(2, 4, 6)

We’ve just lifted the function {(_: Int) * 2} to List[Int] => List[Int]. Here the other derived functions using the operator syntax:

scala> List(1, 2, 3).void
res6: List[Unit] = List((), (), ())

scala> List(1, 2, 3) fproduct {(_: Int) * 2}
res7: List[(Int, Int)] = List((1,2), (2,4), (3,6))

scala> List(1, 2, 3) as "x"
res8: List[String] = List(x, x, x)

Functor Laws 


In order for something to be a functor, it should satisfy some laws. All functors are expected to exhibit certain kinds of functor-like properties and behaviors. … The first functor law states that if we map the id function over a functor, the functor that we get back should be the same as the original functor.

We can check this for Either[A, B].

scala> val x: Either[String, Int] = Right(1)
x: Either[String,Int] = Right(1)

scala> assert { (x map identity) === x }

The second law says that composing two functions and then mapping the resulting function over a functor should be the same as first mapping one function over the functor and then mapping the other one.

In other words,

scala> val f = {(_: Int) * 3}
f: Int => Int = <function1>

scala> val g = {(_: Int) + 1}
g: Int => Int = <function1>

scala> assert { (x map (f map g)) === (x map f map g) }

These are laws the implementer of the functors must abide, and not something the compiler can check for you.