Written by Michael Snoyman | 11/9/16 10:00 AM

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.

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.

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.)

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.

`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.

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.

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.

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 ((() -> () -> ()) -> ())

`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.

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