cvlad

The Functor Family: Contravariant

Prerequisites

This post assumes prior knowledge of

Why

Not all higher kinded types * -> * can have a Functor instance. While types like Maybe a, (x, a), r -> a, Either e a and [a] are Functors in a, there are some types that cannot have a Functor instance. A good example is Predicate:

newtype Predicate a = Predicate { getPredicate :: a -> Bool }

We call this type a predicate in a because, given some value of type a we can obtain a True or a False. So, for example:

We can try writing a Functor instance and see what we can learn:

instance Functor Predicate where
  fmap :: (a -> b) -> Predicate a -> Predicate b
  fmap f (Predicate g) = 
    Predicate
        $ \b -> _welp

As the type hole above would suggest, we need to return a Bool value, and we have:

There is no way we can combine these terms at all, let alone in such a way as to obtain a Bool value. The only way we would be able to obtain a Bool value is by calling g, but for that, we need an a – but all we have is a b.

What if f was reversed, though? If we had f' :: b -> a, then we could apply b to it f' b :: a and then pass it to g to get a Bool. Let’s write this function outside of the Functor class:

mapPredicate :: (b -> a) -> Predicate a -> Predicate b
mapPredicate f (Predicate g) =
    Predicate
        $ \b -> g (f b)

This looks very weird, compared to Functors, right? If you want to go from Predicate a to Predicate b, you need a b -> a function?

Exercise 1: fill in the typed hole _e1:

greaterThan10 :: Predicate Int
greaterThan10 = Predicate (> 10)

exercise1 :: Predicate String
exercise1 = mapPredicate _e1 greaterThan10

Exercise 2: write mapShowable for the following type:

newtype Showable a = Showable { getShowable :: a -> String }

mapShowable :: (b -> a) -> Showable a -> Showable b

Exercise 3: Use mapShowable and showableInt to implement exercise3 such that getShowable exercise3 True is "1" and getShowable exercise3 False is "2".

showableInt :: Showable Int
showableInt = Showable show

exercise3 :: Showable Bool
exercise3 =

How

Predicate and Showable are very similar, and types like them admit a Contravariant instance. Let’s start by looking at it:

class Contravariant f where
  contramap :: (b -> a) -> f a -> f b

The instances for Predicate and Showable are trivial: they are exactly mapPredicate and mapShowable. The difference between Functor and Contravariant is exactly the function they receive: one is “forward” and the other is “backward”, and it’s all about how the data type is defined.

All Functor types have their type parameter a in what we call a positive position. This usually means there can be some a available in the type (which is always the case for tuples, or sometimes the case for Maybe, Either or []). It can also mean we can obtain an a, like is the case for r -> a. Sure, we need some r to do that, but we are able to obtain an a afterwards.

On the opposite side, Contravariant types have their type parameter a in what we call a negative position: they need to consume an a in order to produce something else (a Bool or a String for our examples).

Exercise 4: Look at the following types and decide which can have a Functor instance and which can have a Contravariant instance. Write the instances down:

data T0 a = T0 a Int
data T1 a = T1 (a -> Int)
data T2 a = T2L a | T2R Int
data T3 a = T3
data T4 a = T4L a | T4R a
data T5 a = T5L (a -> Int) | T5R (a -> Bool)

As with Functors, we can partially apply higher kinded types to write a Contravariant instance. The most common case is for the flipped version of ->:

newtype Op a b = Op { getOp :: b -> a }

While a -> b has a Functor instance, because the type is actually (->) a b, and b is in a positive position, its flipped version has a Contravariant instance.

Exercise 5: Write the Contravariant instance for Op:

instance Contravariant (Op r) where
  contramap :: (b -> a) -> Op r a -> Op r b

Exercise 6: Write a Contravariant instance for Comparison:

newtype Comparison a = Comparison { getComparison :: a -> a -> Ordering }

Exercise 7: Can you think of a type that has both Functor and Contravariant instances?

Exercise 8: Can you think of a type that can’t have a Functor nor a Contravariant instance? These types are called Invariant functors.


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