This is the first post in a sequence of write-up about my study of advanced Haskell concepts. As mentioned in the intro to this sequence, this post does not pretend to be a tutorial, merely an exploration of the subject with many explanations.
My first couple of stops on the road to Haskell proficiency
are the most important typeclasses. This includes the infamous
Monad, the unknown Applicative, and the subject of this post:
But before detailing the latter, let's refresh our definition of a
What's a typeclass?
One of Haskell's selling points is its strong type system. We might even say the focus on functional purity stems from the want for powerful type systems. Now, using strong types in a naive way creates massive redundancy: each function would have to be implemented for all the different types it works on. Of course, there will be some implementation to be done -- but the less the better.
Here, genericity comes to the rescue. We use generic types instead of concrete ones to write functions only once -- for the generic type. But this brings another set of problems: with completely generic types, we can't do anything. That's because a type tells us what we can do with its values: you can ask if an int is even or odd, but that doesn't make sense for a bool or a float. For example, a function of type
f :: a -> a
can only be the identity function. Why? Because I don't know anything about
values of type
a. Notably, I don't know how to build one. So when I need
to return a value of type
a, and I got one in input, my only option is
to give it back.
Typeclasses are a clean way to constrain genericity, so that we can implement non trivial functions on generic types. Formally, a typeclass is defined by a set of functions:
class OurTypeclass a where f1 :: a -> [a] f2 :: a -> a -> Bool
An instance of this typeclass for a concrete type is an implementation of these functions.
instance OurTypeclass Bool where f1 x = [x] f2 x y = x
Then this instance gives us the ability to call
f2 on values
Bool. Which means we can define a generic function on any type with
an instance of
OurTypeclass, and use
f2 in this function.
f :: (OurTypeclass a) => a -> a -> [a] f x y = filter (f2 y) (f1 x)
The more types with instances of
OurTypeclass, the most useful
the abstraction becomes.
Another aspect of typeclasses is that instances satsify certain laws. When I say require, the laws are not actually checked at compile-time, since they tend to be undecidable in Haskell's type system. Instead, these laws are promises to Haskell programmers using the generic type. Since programmers write code assuming these promises, instances breaking the laws of their typeclass possibly break any code using this typeclass. So don't do that. Winners don't break typeclass laws.
And lastly, typeclasses generally have a minimal implementation. That is, there is a subset of functions such that all others can be defined in terms of them. So when writing an instance, you only have to implement a set of minimal functions. That being said, you can still reimplement the others; it is mostly done for optimization reasons.
That's it for the typeclass primer. What's left is the one commandment of typeclasses, stressed over and over in every resource that I scoured: laws before burritos. Or more seriously, laws before intuitions. Because it's natural to look for intuitions on what a typeclass captures. Even more when confronted with a weird one like Monad. But by limiting ourselves to these intuitions, we build our understanding on shaky grounds.
Every time I see monads (or other Haskell concepts) explained through intuition alone, I think about Terry Tao's three stages of mathematical education: pre-rigorous, rigorous, and post-rigorous. In summary, we first learn about mathematical concepts in a fuzzy and informal way (the pre-rigorous stage); then we go into the formal details, and rebuild everything from the ground up (the rigorous stage); and finally we can gloss over most details most of the time, since we can stop at any time and unwrap everything (the post-rigorous stage). Learning intuitions without looking at the functions and laws of a typeclass amounts to stagnating at the pre-rigorous stage of Haskell education. And since I aim for post-rigorous mastery, the only way is the rigorous path.
A first look at Functor
Form and functions
class Functor f where fmap :: (a -> b) -> f a -> f b (<$) :: a -> f b -> f a (<$) = fmap . const
First, one subtlety I struggled with: instances of
functions on types. That is, a type that takes
another type and gives a concrete type. A simple example is the list type
 a is actually equivalent to
 has an instance of
Functor, but not
Technically, this means that the kind (the "type of a type") of a
* -> *, where
* is the kind of a concrete datatype like
Bool or even
Carrying on, the meat of this class is the
fmap function. Indeed, it constitutes a minimal
Functor: the other function
(<$) follows by default from
the definition of
fmap. As a first step, let's look at the type of
fmap :: (a -> b) -> f a -> f b
fmap takes a function of type
a -> b, a value of type
f a, and returns
a value of type
f b. We get a transformation from one type to another, an element
of our functor applied to the first type, and must transform it into an element of
our functor applied to the second type.
If you're able to write such a function for your type
f, then you have a
The other function is just a specialization of
fmap when the transformation is
constant. So instead of giving a function, we just give an element of the return
type of the function for
fmap. (Also notice the parentheses, telling us
this function use infix notation,
like the + operator).
(<$) :: a -> f b -> f a
What confused me at first is that
b are reversed
fmap. But that's only the consequence of Haskell's naming
convention for types: the first type variable of kind * is named
b, the third
c... If I break this convention and use the
same names than for
fmap, I get a type fitting the description above:
(<$) :: b -> f a -> f b
As I mentioned,
(<$) has a default implementation in terms of
(<$) = fmap . const
And given that the only reason to rewrite a default implementation, and
that all the work is done by
fmap, I see no reason for ever writing
(<$). That said, I'll be pretty interested by an counter-example
if one exists.
Finally for the functions, there are utility functions defined on
Theses cannot be reimplemented, as they are not part of the typeclass.
(<$>) :: Functor f => (a -> b) -> f a -> f b (<&>) :: Functor f => f a -> (a -> b) -> f b ($>) :: Functor f => f a -> b -> f b void :: Functor f => f a -> f ()
These are not that interesting:
(<$>) is the infix version of
($>) are respectively the flipped versions of
void just replaces the inside of a functor by the unit value
the unit type.
I lied when I said that any function of type
(a -> b) -> f a -> f b gives
fmap, and an acceptable instance of
Functor. In fact,
instances must follow the laws:
fmap id = id fmap (g . f) = (fmap g) . (fmap f)
As mentioned above, these laws are not enforced by GHC. My first instinct was that they were undecidable. After all, function equality is undecidable. But after looking a bit more into the matter, the only thing I'm sure of is that GHC can't do it. On the other hand, adding flavors of dependent types to the mix allow semi-automatic verification of these laws. Although you and I will probably never use these, or at least not in the near future.
So, these laws are not checked at compile-time. And breaking them doesn't (in general)
create runtime panics and exceptions. Nonetheless, they matter: they
capture what you can expect from a
Functor instance. That is to say, programmers
using your type as a
fmap mostly) will expect it to satisfy these
laws. And their code might behave incorrectly if these assumptions
Going back to the laws, they clarify a little more what a
is. The first one,
fmap id = id
says that when passing the identity function to
fmap, it does nothing. This
fits the intuition that a
Functor is some sort of container, a structure
with values of type
a inside. And
fmap can only touch these values,
not the structure itself.
Let's look at the example:
map, the instance of
fmap for lists. It applies the function
to every element of the list, but doesn't change the number of elements, or
As for the second law, it constrains
Functors even more towards containers:
fmap (g . f) = (fmap g) . (fmap f)
When applying a function
fmap and then a function
fmap, that must be equivalent to applying the composition
g . f through
This also holds for
map, since applying
g to each element
is the same as applying
g . f to each element.
Examples and anti-examples
When I learn a definition, I like to get both examples and anti-examples (examples not satisfying the definition). That gives me an outline of the concept from the inside (what it requires) and from the outside (what it forbids).
Let's start with the instance for list, which I already mentioned:
instance Functor  where fmap = map
As explained above, this instance satisfies the
Functor laws. Can we write
one that doesn't? That's pretty easy.
instance Functor  where fmap f l = 
instance Functor  where fmap f  =  fmap f x:xs = (fmap f xs)++[f x]
Now, these two instances break both laws. Can we break one and not the other? We'll see later that the second law follows from the first law; so we can't break the second without breaking the first. But can we break the first without breaking the second?
If the second law is true, then
fmap (g . id) = (fmap g) . (fmap id) fmap g = (fmap g) . (fmap id)
fmap (id . g) = (fmap id) . (fmap g) fmap g = (fmap id) . (fmap g)
This entails that
fmap id acts like the identity on elements of
f b depending on the side. I think this means that the second law
implies the first law, but it's slightly tricky. This is because not
all functions of type
f a -> f b are representable as
fmap g for
g :: a -> b. For example, the function applying
g on each
element of a list and reversing the list is not representable by an
fmap g = (fmap g) . (fmap id) is not
h = h . (fmap id), and similarly for the other direction.
Thus I think the second law implies the first law, but I would appreciate any clean proof or counter-example.
Maybe. The canonical instance is
instance Functor Maybe where fmap f Nothing = Nothing fmap f (Just x) = Just (f x)
Here too, we can write blatantly false instances.
instance Functor Maybe where fmap f x = Nothing
and... that's it actually. I just realized while coming up with examples
Maybe only has this bad
Functor instance. Because with
you can only return
Nothing (nothing to apply
f on), and with
the only way to get a value of type
b from it is by applying
The other option being returning
The intuition behind the
Maybe is that it applies the function
when there is something, and it keeps
Nothing when there isn't. Pretty
One last example:
Either. The thing that disturbed me with
that its kind is actually
* -> * -> *, because it takes the type
Left (the error type) and the type of the
Right (the result type).
Which means that for getting a
Functor, we need to provide the first parameter
(the error type).
instance Functor (Either e) where fmap f (Left e) = (Left e) fmap f (Right x) = (Right f x)
Either acts like the one for
Maybe, except that it keeps
the error if there was one before.
Here, there are no false instance, because we cannot change what
we do to
Right x, and we have no way of building a value of type
You might have noticed that all the examples I gave have lawful instances of
Is this always the case? No: one type without a
Functor instance is the following.
data Type a = Type (a -> Bool)
Why? Because there is no way to build a value of
b -> Bool from
a value of
a -> Bool and a value of
a -> b. The last two types
are not composable, so we're screwed.
How do we distinguish types with a
Functor and types without?
Glad you asked; let's go down the first rabbit hole.
Down the rabbit holes
Existence: covariance and contravariance
The example above, of a type without a
Functor, I did not find myself.
Instead, after failing to find such an example, I asked Google and found
it on stack overflow.
I convinced myself that it has no implementation for
fmap. But what is the
underlying reason? How am I supposed to know if a given type has a
Apparently, types with a
Functor instance must be covariant according
to their type parameter. And our counter-example above is contravariant,
which means it has no
Functor. So that's just two new words. What do they
mean? When is a type covariant according to a type parameter, and what does
it means concretely?
First, there is a typeclass with the contravariant equivalent of
class Contravariant f where contramap :: (a -> b) -> f b -> f a (>$) :: b -> f b -> f a (>$) = contramap . const
Type above has an instance of
instance Contravariant Type where contramap f (Type g) = Type (g . f)
The difference is that the function given to contramap goes the other
way. And from a function
a -> Bool and a function
b -> a, composition
gives a function
b -> Bool.
The only way to have both an instance of
Functor and of
Contravariant is for
the type parameter to be phantom, that is to say useless:
data Useless a = Useless instance Functor Useless where fmap _ Useless = Useless instance Contravariant Useless where contramap _ Useless = Useless
And some types have neither an instance of
Functor nor one of
data Nope a = Nope (a -> a)
So we have four cases. Looking around, it seems the type parameter can be in one of four relations with the type itself, which entails in which case it lies:
ais covariant, then it has a
Functorinstance and no
ais contravariant, then it has no
Functorinstance and a
ais bivariant, then it has no
Functorinstance and no
ais invariant, then it has both a
Functorinstance and a
That still does not answer my question about the meaning of covariant and contravariant. For that, the best source I found, surprisingly, was the wikipedia page.
Basically, what matters is the relation between subtypes of the type parameters,
and subtypes of the full type. The most interesting example is the function (at
least in a purely functional context like Haskell). Let's say I have a function
a -> b, where
b are fixed types. What are the subtypes
a -> b? That is, what are the types of objects with which I can replace
f, and still have correct types?
f takes a value of type
a. This means that when I use it in
my code, I will pass to it a value of this type. Now assume I have a
A that is a supertype of
a. For example,
A is a generic
a is one of its concrete types. In Haskell,
A would be
the generic type associated with a typeclass for which
a has an instance.
Then I can replace
f by a function of type
A -> b, and the program will
still compile. On the other hand, if
a is a generic type associated with
a typeclass, and, let's say
Bool has an instance of this typeclass,
I cannot replace
f by a function of type
Bool -> b. Because the
f entails that it can receive any value of type
not just values from
This whole reasoning is reversed for
f returns a value
b, my program will expect something of this type. Which means
that if we want to change the return type, then it must be a subtype
b, not a supertype.
Why do we care? Well, because this gives us the definition of covariant
and contravariant. A subtype of
a -> b returns a subtype of
the type parameter and the full type vary in the same direction. We say
a -> b is covariant in
b. Whereas a subtype of
a -> b
takes a supertype of
a; this means, you guessed it, that
a -> b
is contravariant in
This gives us a way to check, given a type and its type parameter, whether the former is covariant/contravariant/bivariant/invariant in the latter. For functions, which are pretty much the only place where contravariant positions happens, the wikipedia page gives a rule of thumbs.
a position is covariant if it is on the left side of an even number of arrows applying to it.
Easy, no? Well, it took me a discussion with a colleague expert in type theory to realize I misread this. It does not say "if it is on the left side of an even number of arrows", but "if it is on the left side of an even number of arrows applying to it". Why the fuss? Because this type
data Tricky a = Tricky (a -> (Bool -> Int))
has an even number of arrows on the right of
a, but it is contravariant in
This is because the last arrow doesn't apply to
a -- it is on the other side
of the principal arrow, the root of the binary tree capturing the type expression.
On the other hand, the type
data NotTricky a = NotTricky ((a -> Bool) -> Int)
is covariant in
instance Functor NotTricky where fmap f (NotTricky h) = (\g -> h (g . f))
Because from a
a -> b and a
b -> Bool, I can build a
a -> Bool.
And giving that to
((a -> Bool) -> Int), I get a
Int. So from
f:: a -> b and an element of
NotTricky a, I can build an element of
So, caveat aside, we have a rule for computing whether a type is covariant
or contravariant, or both or neither, in a type parameter. That's almost
what I was looking for. What is still missing is an understanding of
fmap requires a type that is covariant in its type parameter. I mean,
I can check for the examples, like I did above. But I did not find a general
Yet this is enough for my purpose. That being said, if someone has such a reference or explanation, I'll gladly take it.
Unicity: Free Theorems!
Once settled, somewhat, the question of whether a
Functor exists for
a given type, a natural follow-up is whether there are multiple instances
for this type. As a matter of fact, there is only one instance of
(satisfying the laws) per type. I read that in the Typeclassopedia,
but when I followed the links...
let's just say that Haskell is full of rabbit holes.
The proof of the unicity of
Functor instances follows from the free theorem for
the type of
fmap. What is a free theorem? Well... here goes the rabbit hole.
The opening in the earth is this
paper by Wadler,
titled "Theorems for free!". In it, he shows
that given a polymorphic type, we can deduce a theorem for all values of this
type. That is, just from the type, we can deduce things about any instance of it.
Remember when I said that the only function of type
a -> a is the identity? That
actually follows from the free theorem for
a -> a : if
f :: a -> a, then
g, we have
g . f = f . g. The only
f satisfying this property is
the identity function (with the usual caveat that only the input/output behavior
matters here, so we work up to isomorphism).
So the free theorem gives a property that any function of this type must satisfy.
The way we show the unicity of the
Functor instance is by assuming the existence
of one instance satisfying the functor laws. Let's
fmap be the function
defined for this instance. Then, if we have
foo :: (a -> b) -> f a -> f b satisfying the functor laws for our type, we
use the free theorem on
foo. It is: for
f,g,h,k such that
g . k = h . f
fmap g . foo k = foo h . fmap f
Free theorems show their powers when specialized. Here, we can choose
g = h and
f = k = id, which maintains the condition
g . k = g = f = h . f.
Then the free theorem specializes into
fmap g . foo id = foo g . fmap id
which by the first law of functors (satisfied by both
foo) gives us:
fmap g = foo g
That is, foo has exactly the same input output behavior than
A similar reasoning
can also be used to show that
fmap only has to satisfy the first law
-- the second law follows from the first and the free theorem for
All of this is actually fascinating, but I don't want to get into the derivations of free theorems in this post. First because it is already long enough; and second because free theorems deserve a post of their own.
Automation: Deriving Functor Instances
The nice thing about having only one instance of
Functor, if it exists,
is that we can derive it automatically: the
of GHC does exactly that.
Although the page I linked is quite thorough, I didn't find much to learn in it.
Almost all of it seems to talk about when a
Functor instance cannot be derived...
that is when the type is contravariant or bivariant in its
last type parameter. So apparently, GHC
cannot derive a
Functor instance iff there is no such instance. That's pretty
Now that we did our homework on the technical details, we can go through some
common intuitions about
Functor. I think we deserved it.
Is it a container? Is it a context? No, it's a Functor!
First, there are two "obvious" interpretation of a
Functor in Haskell:
as a container, and as a context.
Functors can be thought
as either, some works really well as one or the other. For example, lists
are very natural containers. A list is a sequence of values, and applying
something on this container translate into applying it to its values.
On the other hand,
Either works well as a context. It captures
whether everything is alright, or whether an error happened.
Most examples I have in mind fit in one of these categories. But that's not the case of all functors:
data SomeType a = SomeType (Bool -> a)
represents neither a container of values of type
a, nor a context
a is embedded. It is more of a producer of
a. And I
don't even have an intuition like that for the weirder function
types where a is covariant, like
(a -> Bool) -> Int.
Let that remind us that intuitions only go so far.
Going to the next level
Another perspective on
Functor appears if we change parentheses slightly:
fmap :: (a -> b) -> (f a -> f b)
That is, instead of seeing
fmap as taking a function and a functor, and returning
a transformed functor, we can see it as taking a function and returning it
on the functors of the types. The jargon for this is a lift. From discussions
with my friends about Haskell, and category theory, and types, I know there's
a lot coming on the topic of lifts. But for
now, it's only a nice new perspective on
Category Theory is watching you
Finally, let's address the elephant in the room: the name
from category theory. And yet, I did not mention it during this post.
Right now, I'm not sure what it teaches me. I mean, I get that functors
are mapping from one category to another, and that they maintain identity
and composition in the same way that the
Functor laws do.
I even get that a Haskell
Functor is a functor on the category
Hask of Haskell types (with some subtleties).
Nonetheless, I cannot see, for now, a use for this understanding. I'll keep searching, but I think that the connection brings fruits only for concepts more complex than mere functors.
On the other hand, going deeper into typeclasses made me think about the links between Haskell and category theory. I feel that the implementation of powerful typeclasses is akin to the effort in category theory for finding the right category and transformations to do the job. Then in one case you have a short generic program, and in the other you have a small generic diagram. And in both cases, when you want to understand exactly what is happening, you need to go down into the nitty gritty details hidden behind the cute abstractions: instances in Haskell and the meaning of objects and arrows in category theory.
There sure are a lot of things to dig in Haskell!
I came in pretty sure that I knew what
Functor was, and I still learned