Providence Salumu Covariance and Contravariance :: FP Complete

Covariance and Contravariance

9 Nov 2016 Michael Snoyman

Typeclasses such as Bifunctor are often expressed in terms of whether they are covariant or contravariant. While these terms may appear intimidating to the unfamiliar, they are a precise language for discussing these concepts, and once explained are relatively easy to understand. Furthermore, the related topics of positive and negative position can greatly simplify how you think about complex data structures. This topic also naturally leads into subtyping.

This post is intended to give a developer-focused explanation of the terms without diving into the category theory behind them too much. For more information, please see the Wikipedia page on covariance and contravariance.

This blog post is also part of the FP Complete Haskell Syllabus and part of our Haskell training.

The Functor typeclass: covariant functor

Let's consider the following functions (made monomorphic for clarity):

showInt :: Int -> String
showInt = show

floorInt :: Double -> Int
floorInt = floor

Now suppose that we have a value:

maybeInt :: Maybe Int
maybeInt = Just 5

We know Maybe is an instance of Functor, providing us with the following function:

fmapMaybe :: (a -> b) -> Maybe a -> Maybe b
fmapMaybe = fmap

We can use fmapMaybe and showInt together to get a new, valid, well-typed value:

maybeString :: Maybe String
maybeString = fmapMaybe showInt maybeInt

However, we can't do the same thing with floorInt. The reason for this is relatively straightforward: in order to use fmapMaybe on our Maybe Int, we need to provide a function that takes an Int as an input, whereas floorInt returns an Int as an output. This is a long-winded way of saying that Maybe is covariant on its type argument, or that the Functor typeclass is a covariant functor.

Doesn't make sense yet? Don't worry, it shouldn't. In order to understand this better, let's contrast it with something different.

A non-covariant data type

Consider the following data structure representing how to create a String from something:

newtype MakeString a = MakeString { makeString :: a -> String }

We can use this to convert an Int into a String:

newtype MakeString a = MakeString { makeString :: a -> String }

showInt :: MakeString Int
showInt = MakeString show

main :: IO ()
main = putStrLn $ makeString showInt 5

The output for this program is, as expected, 5. But suppose we want to both add 3 to the Int and turn it into a String. We can do:

newtype MakeString a = MakeString { makeString :: a -> String }

plus3ShowInt :: MakeString Int
plus3ShowInt = MakeString (show . (+ 3))

main :: IO ()
main = putStrLn $ makeString plus3ShowInt 5

But this approach is quite non-compositional. We'd ideally like to be able to just apply more functions to this data structure. Let's first write that up without any typeclasses:

newtype MakeString a = MakeString { makeString :: a -> String }

mapMakeString :: (b -> a) -> MakeString a -> MakeString b
mapMakeString f (MakeString g) = MakeString (g . f)

showInt :: MakeString Int
showInt = MakeString show

plus3ShowInt :: MakeString Int
plus3ShowInt = mapMakeString (+ 3) showInt

main :: IO ()
main = putStrLn $ makeString plus3ShowInt 5

But this kind of mapping inside a data structure is exactly what we use the Functor type class for, right? So let's try to write an instance!

instance Functor MakeString where
    fmap f (MakeString g) = MakeString (g . f)

Unfortunately, this doesn't work:

Main.hs:4:45:
    Couldn't match type ‘b’ with ‘a’
      ‘b’ is a rigid type variable bound by
          the type signature for
            fmap :: (a -> b) -> MakeString a -> MakeString b
          at Main.hs:4:5
      ‘a’ is a rigid type variable bound by
          the type signature for
            fmap :: (a -> b) -> MakeString a -> MakeString b
          at Main.hs:4:5
    Expected type: b -> a
      Actual type: a -> b
    Relevant bindings include
      g :: a -> String (bound at Main.hs:4:24)
      f :: a -> b (bound at Main.hs:4:10)
      fmap :: (a -> b) -> MakeString a -> MakeString b
        (bound at Main.hs:4:5)
    In the second argument of ‘(.)’, namely ‘f’
    In the first argument of ‘MakeString’, namely ‘(g . f)’

To understand why, let's compare the type for fmap (specialized to MakeString) with our mapMakeString type:

mapMakeString :: (b -> a) -> MakeString a -> MakeString b
fmap          :: (a -> b) -> MakeString a -> MakeString b

Notice that fmap has the usual a -> b parameter, whereas mapMakeString instead has a b -> a, which goes in the opposite direction. More on that next.

Exercise: Convince yourself that the mapMakeString function has the only valid type signature we could apply to it, and that the implementation is the only valid implementation of that signature. (It's true that you can change the variable names around to cheat and make the first parameter a -> b, but then you'd also have to modify the rest of the type signature.)

Contravariance

What we just saw is that fmap takes a function from a -> b, and lifts it to f a -> f b. Notice that the a is always the "input" in both cases, whereas the b is the "output" in both cases. By contrast, mapMakeString has the normal f a -> f b, but the initial function has its types reversed: b -> a. This is the core of covariance vs contravariance:

  • In covariance, both the original and lifted functions point in the same direction (from a to b)
  • In contravariance, the original and lifted functions point in opposite directions (one goes from a to b, the other from b to a)

This is what is meant when we refer to the normal Functor typeclass in Haskell as a covariant functor. And as you can probably guess, we can just as easily define a contravariant functor. In fact, it exists in the contravariant package. Let's go ahead and use that typeclass in our toy example:

import Data.Functor.Contravariant

newtype MakeString a = MakeString { makeString :: a -> String }

instance Contravariant MakeString where
    contramap f (MakeString g) = MakeString (g . f)

showInt :: MakeString Int
showInt = MakeString show

plus3ShowInt :: MakeString Int
plus3ShowInt = contramap (+ 3) showInt

main :: IO ()
main = putStrLn $ makeString plus3ShowInt 5

Our implementation of contramap is identical to the mapMakeString used before, which hopefully isn't too surprising.

Example: filtering with Predicate

Let's say we want to print out all of the numbers from 1 to 10, where the English word for that number is more than three characters long. Using a simple helper function english :: Int -> String and filter, this is pretty simple:

greaterThanThree :: Int -> Bool
greaterThanThree = (> 3)

lengthGTThree :: [a] -> Bool
lengthGTThree = greaterThanThree . length

englishGTThree :: Int -> Bool
englishGTThree = lengthGTThree . english

english :: Int -> String
english 1 = "one"
english 2 = "two"
english 3 = "three"
english 4 = "four"
english 5 = "five"
english 6 = "six"
english 7 = "seven"
english 8 = "eight"
english 9 = "nine"
english 10 = "ten"

main :: IO ()
main = print $ filter englishGTThree [1..10]

The contravariant package provides a newtype wrapper around such a -> Bool functions, called Predicate. We can use this newtype to wrap up our helper functions and avoid explicit function composition:

import Data.Functor.Contravariant

greaterThanThree :: Predicate Int
greaterThanThree = Predicate (> 3)

lengthGTThree :: Predicate [a]
lengthGTThree = contramap length greaterThanThree

englishGTThree :: Predicate Int
englishGTThree = contramap english lengthGTThree

english :: Int -> String
english 1 = "one"
english 2 = "two"
english 3 = "three"
english 4 = "four"
english 5 = "five"
english 6 = "six"
english 7 = "seven"
english 8 = "eight"
english 9 = "nine"
english 10 = "ten"

main :: IO ()
main = print $ filter (getPredicate englishGTThree) [1..10]

NOTE: I'm not actually recommending this as a better practice than the original, simpler version. This is just to demonstrate the capability of the abstraction.

Bifunctor and Profunctor

We're now ready to look at something a bit more complicated. Consider the following two typeclasses: Profunctor and Bifunctor. Both of these typeclasses apply to types of kind * -> * -> *, also known as "a type constructor that takes two arguments." But let's look at their (simplified) definitions:

class Bifunctor p where
    bimap :: (a -> b) -> (c -> d) -> p a c -> p b d

class Profunctor p where
    dimap :: (b -> a) -> (c -> d) -> p a c -> p b d

They're identical, except that bimap takes a first parameter of type a -> b, whereas dimap takes a first parameter of type b -> a. Based on this observation, and what we've learned previously, we can now understand the documentation for these two typeclasses:

Bifunctor: Intuitively it is a bifunctor where both the first and second arguments are covariant.

Profunctor: Intuitively it is a bifunctor where the first argument is contravariant and the second argument is covariant.

These are both bifunctors since they take two type parameters. They both treat their second parameter in the same way: covariantly. However, the first parameter is treated differently by the two: Bifunctor is covariant, and Profunctor is contravariant.

Exercise Try to think of a few common datatypes in Haskell that would be either a Bifunctor or Profunctor, and write the instance.

Hint Some examples are Either, (,), and -> (a normal function from a to b). Figure out which is a Bifunctor and which is a Profunctor.

Solution

class Bifunctor p where
    bimap :: (a -> b) -> (c -> d) -> p a c -> p b d

class Profunctor p where
    dimap :: (b -> a) -> (c -> d) -> p a c -> p b d


instance Bifunctor Either where
    bimap f _ (Left x) = Left (f x)
    bimap _ f (Right x) = Right (f x)
instance Bifunctor (,) where
    bimap f g (x, y) = (f x, g y)

instance Profunctor (->) where -- functions
    dimap f g h = g . h . f

Make sure you understand why these instances work the way they do before moving on.

Bivariant and invariant

There are two more special cases for variance: bivariant means "both covariant and contravariant," whereas invariant means "neither covariant nor contravariant." The only types which can be bivariant are phantoms, where the type doesn't actually exist. As an example:

import Data.Functor.Contravariant (Contravariant (..))

data Phantom a = Phantom
instance Functor Phantom where
    fmap _ Phantom = Phantom
instance Contravariant Phantom where
    contramap _ Phantom = Phantom

Invariance will occur if:

  • A type parameter is used multiple times in a data structure, both positively and negatively, e.g.:

    data ToFrom a = ToFrom (a -> Int) (Int -> a)

  • A type parameter is used in type which is itself invariant in the parameter, e.g.:

    newtype ToFromWrapper a = ToFromWrapper (ToFrom a)

    Note that even though the parameter only appears once here, it appears twice in ToFrom itself.

  • In special types like references, e.g.:

    data IORef a -- a is invariant
newtype RefWrapper a = RefWrapper (IORef a) -- also invariant

Exercise Convince yourself that you can not make an instance of either Functor nor Contravariant for ToFrom or IORef.

Exercise Explain why there's also no way to make an instance of Bifunctor or Profunctor for these datatypes.

As you can see, the a parameter is used as both the input to a function and output from a function in ToFrom. This leads directly to our next set of terms.

NOTE There's a good Reddit discussion which led to clarification of these section.

Positive and negative position

Let's look at some basic covariant and contravariant data types:

data WithInt a = WithInt (Int -> a)
data MakeInt a = MakeInt (a -> Int)

By now, you should hopefully be able to identify that WithInt is covariant on its type parameter a, whereas MakeInt is contravariant. Please make sure you're confident of that fact, and that you know what the relevant Functor and Contravariant instance will be.

Can we give a simple explanation of why each of these is covariant and contravariant? Fortunately, yes: it has to do with the position the type variable appears in the function. In fact, we can even get GHC to tell us this by using Functor deriving:

{-# LANGUAGE DeriveFunctor #-}

data MakeInt a = MakeInt (a -> Int)
    deriving Functor

This results in the (actually quite readable) error message:

Can't make a derived instance of ‘Functor MakeInt’:
  Constructor ‘MakeInt’ must not use the type variable in a function argument
In the data declaration for ‘MakeInt’

Another way to say this is "a appears as an input to the function." An even better way to say this is that "a appears in negative position." And now we get to define two new terms:

  • Positive position: the type variable is the result/output/range/codomain of the function
  • Negative position: the type variable is the argument/input/domain of the function

When a type variable appears in positive position, the data type is covariant with that variable. When the variable appears in negative position, the data type is contravariant with that variable. To convince yourself that this is true, go review the various data types we've used above, and see if this logic applies.

But why use the terms positive and negative? This is where things get quite powerful, and drastically simplify your life. Consider the following newtype wrapper intended for running callbacks:

type Callback a = a -> IO ()
-- newtype CallbackRunner a = CallbackRunner (Callback a -> IO ())
-- Expands to:
newtype CallbackRunner a = CallbackRunner ((a -> IO ()) -> IO ())

Is it covariant or contravariant on a? Your first instinct may be to say "well, a is a function parameter, and therefore it's contravariant. However, let's break things down a bit further.

Suppose we're just trying to deal with a -> IO (). As we've established many times above: this function is contravariant on a, and equivalently a is in negative position. This means that this function expects on input of type a.

But now, we wrap up this entire function as the input to a new function, via: (a -> IO ()) -> IO (). As a whole, does this function consume an a, or does it produce an a? To get an intuition, let's look at an implementation of CallbackRunner Int for random numbers:

supplyRandom :: CallbackRunner Int
supplyRandom = CallbackRunner $ \callback -> do
    int <- randomRIO (1, 10)
    callback int

It's clear from this implementation that supplyRandom is, in fact, producing an Int. This is similar to Maybe, meaning we have a solid argument for this also being covariant. So let's go back to our positive/negative terminology and see if it explains why.

In a -> IO (), a is in negative position. In (a -> IO ()) -> IO (), a -> IO () is in negative position. Now we just follow multiplication rules: when you multiply two negatives, you get a positive. As a result, in (a -> IO ()) -> IO (), a is in positive position, meaning that CallbackRunner is covariant on a, and we can define a Functor instance. And in fact, GHC agrees with us:

{-# LANGUAGE DeriveFunctor #-}
import System.Random

newtype CallbackRunner a = CallbackRunner
    { runCallback :: (a -> IO ()) -> IO ()
    }
    deriving Functor

supplyRandom :: CallbackRunner Int
supplyRandom = CallbackRunner $ \callback -> do
    int <- randomRIO (1, 10)
    callback int

main :: IO ()
main = runCallback supplyRandom print

Let's unwrap the magic, though, and define our Functor instance explicitly:

newtype CallbackRunner a = CallbackRunner
    { runCallback :: (a -> IO ()) -> IO ()
    }

instance Functor CallbackRunner where
    fmap f (CallbackRunner aCallbackRunner) =
        CallbackRunner $ \bCallback ->
            aCallbackRunner (bCallback . f)

Exercise 1: Analyze the above Functor instance and understand what is occurring.

Exercise 2: Convince yourself that the above implementation is the only one that makes sense, and similarly that there is no valid Contravariant instance.

Exercise 3: For each of the following newtype wrappers, determine if they are covariant or contravariant in their arguments:

newtype E1 a = E1 (a -> ())
newtype E2 a = E2 (a -> () -> ())
newtype E3 a = E3 ((a -> ()) -> ())
newtype E4 a = E4 ((a -> () -> ()) -> ())
newtype E5 a = E5 ((() -> () -> a) -> ())

-- trickier:
newtype E6 a = E6 ((() -> a -> a) -> ())
newtype E7 a = E7 ((() -> () -> a) -> a)
newtype E8 a = E8 ((() -> a -> ()) -> a)
newtype E9 a = E8 ((() -> () -> ()) -> ())

Lifting IO to MonadIO

Let's look at something seemingly unrelated to get a feel for the power of our new analysis tools. Consider the base function openFile:

openFile :: FilePath -> IOMode -> IO Handle

We may want to use this from a monad transformer stack based on top of the IO monad. The standard approach to that is to use the MonadIO typeclass as a constraint, and its liftIO function. This is all rather straightforward:

import System.IO
import Control.Monad.IO.Class

openFileLifted :: MonadIO m => FilePath -> IOMode -> m Handle
openFileLifted fp mode = liftIO (openFile fp mode)

But of course, we all prefer using the withFile function instead of openFile to ensure resources are cleaned up in the presence of exceptions. As a reminder, that function has a type signature:

withFile :: FilePath -> IOMode -> (Handle -> IO a) -> IO a

So can we somehow write our lifted version with type signature:

withFileLifted :: MonadIO m => FilePath -> IOMode -> (Handle -> m a) -> m a

Try as we might, this can't be done, at least not directly (if you're really curious, see lifted-base and its implementation of bracket). And now, we have the vocabulary to explain this succinctly: the IO type appears in both positive and negative position in withFile's type signature. By contrast, with openFile, IO appears exclusively in positive position, meaning our transformation function (liftIO) can be applied to it.

Like what you learned here? Please check out the rest of our Haskell Syllabus or learn about FP Complete training.

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