9 Jan 2023
6 min read
This is the second of a series of articles that deep dives a few very powerful type classes that every functional programmer should know. For example,
Monad are all prevalent in production code. We explain them with examples in
Haskell. This series assumes the audience is familiar with some functional programming concepts we went through in the previous series.
In the previous post, we went through one of the most universal typeclasses:
Semigroup. In this post, we go through a subclass of
Monoid is also very prevalent and we will look at some notable monoids.
Recall that a typeclass definition consists of:
The typeclasses in functional programming are based on abstract algebra. In abstract algebra, a
Monoid is a set equipped with an associative binary operation and an identity element.
In Haskell, only a type that has a
Semigroup instance can have a
class Semigroup a => Monoid a where
A type has a
Monoid instance if it has the following methods:
Semigroup) an associative binary function (
<>); sometimes called its obsolete/deprecated name
<> is the same as the
Semigroup method. But
Monoid has an additional method (
Monoid is a subset of
Semigroup, or, to say it the other way around,
Semigroup is a superset of
Monoid. This means that every
Monoid is a
Semigroup, but not every
Semigroup is a
The methods has to satisfy the following laws:
For the binary function: as we know from the
Semigroup law, it has to be associative. I.e.:
x <> (y <> z) = (x <> y) <> z
For the identity element: anything
<> to the identity element has to return itself.
Right identity: anything
mempty on the right) is itself
x <> mempty = x
Left identity: same but with
mempty to the left of
mempty <> x = x
Monoid class also provides a default concatenation function. It has the following definition:
mconcat = foldr (<>) mempty
mconcat is an extra method based on them. So there is no new method required technically. This is an extra function provided by the
Monoid library in Haskell that you can use by default.
Looking at the signature and definition of
foldr :: (a -> b -> b) -> b -> [a] -> b
foldr applied to the binary operator, with the starting value of the identity element. So its signature is:
mconcat :: [a] -> b
mconcat takes a list and
<> the list from right to left. We will see some examples of this below.
In the previous post, we reviewed a few common
Semigroup instances. Some of them are also
Monoid instances, but some of them aren't.
Let's look at every typeclass we went through previous time and try it in GHCi.
From the previous post we know that the
<> function for
<> is the same as
mempty, because the empty list
 has the property of the identity element for list, it's the same as
mempty for list. Trying it out in GHCi:
Prelude> <>[1,2,3] [1,2,3] Prelude> [1,2,3]<>mempty [1,2,3] Prelude> (mempty :: [a]) -- if we specify that `mempty` is of type list, it returns an empty list 
mconcat, let's look at the type signature of
foldr :: (a -> b -> b) -> b -> [a] -> b
<> function takes in lists, and return a list, so
b above are of type
List here. And we need to give
mconcat a list of lists.
<> the list of lists from right to left, returning a
List. This is the same as the function
In fact, this is a common use of
Prelude> mconcat [[1,2,3],,[4,5,6],,,[8,9],,] [1,2,3,4,5,6,7,8,9,0]
OK, this all makes sense and list is definitely a
Recall that the
NonEmpty data family is a list that has at least one element in it. So
mempty is not defined in
NonEmpty data family is not an instance of
Monoid! GHCi confirms it:
Prelude> import Data.List.NonEmpty -- GHCi complains about not having a `Monoid` instance for `NonEmpty Int` Prelude Data.List.NonEmpty> (mempty :: NonEmpty Int) <interactive>:2:2: error: • No instance for (Monoid (NonEmpty Int)) arising from a use of ‘mempty’ • In the expression: (mempty :: NonEmpty Int) In an equation for ‘it’: it = (mempty :: NonEmpty Int)
From the previous post we know how the
<> function works for
Maybe a. Recall that the type variable
a has to have a
Semigroup instance for the
<> function to work:
Semigroup a => Semigroup (Maybe a) -- for `<>` we only need Semigroup
Maybe is listed as an instance of
Monoid if the type variable has a
Semigroup a => Monoid (Maybe a)
Either Integer String is a
Semigroup instance, so
Maybe (Either Integer String) has a
Monoid instance. Let's try it out in GHCi:
Prelude> import Data.Maybe Prelude Data.Maybe> Just (Right 1) <> Just (Left "string") Just (Right 1) -- `Nothing` satisfies the laws for the identity element. Prelude Data.Maybe> Just (Left "a left") <> Nothing Just (Left "a left") Prelude Data.Maybe> Nothing <> Just (Right 2) Just (Right 2) -- `mempty` and `Nothing` are the same for the `Maybe` typeclass. Prelude Data.Maybe> Just (Left "a left") <> mempty Just (Left "a left") Prelude Data.Maybe> mempty <> Just (Right 2) Just (Right 2)
mconcat, recall that it takes in a list of something, and returns a something. Here we have
Maybe (Either Integer String) as the something. Let's give it a list of that, and we should get just a
Maybe (Either Integer String) back:
Prelude Data.Maybe> mconcat [Just (Right 1), mempty, Just (Left "a left"), mempty] Just (Right 1)
Recall from the previous post that association of
Maybe a works by applying
a's while ignoring
Nothing is the identity element!). That's why we need
a to have a
Semigroup instance. So what's happening is:
mconcat [Just (Right 1), mempty, Just (Left "a left"), mempty] -- recall that `mconcat` is `foldr` with some inputs, so it works from right to left. => Just (Right 1) <> (mempty <> (Just (Left "a left") <> mempty)) => Just (Right 1) <> (mempty <> Just (Left "a left")) => Just (Right 1) <> Just (Left "a left") => Just (Right 1 <> Left "a left") -- `<>` for `Either` returns the first `Right` and if there is no `Right`, returns the last `Left`. => Just (Right 1)
It's interesting to note that because
Maybe essentially lets us say it's just
mempty), it is the simplest way to give anything with a
Semigroup instance a
Looking at the documentation, you won't be able to find a
Monoid instance for
Either! This is because there is no identity element for
Either. Recall that
Either only has two constructors:
data Either a b = Left a | Right b
This means the empty element has to be either
Left a or
Right b! We cannot construct it any other way!
We know that
Either returns the first
Right definitely cannot be the identity element since it violates the left identity law (
mempty <> x = x).
Also, we know that if there is no
<> returns the last
Left cannot be the identity element either, because it violates the right identity law (
x <> mempty = x).
The requirements of
Monoid seem simple, but it's because it's so simple that the concept of
Monoid is very useful!
In an earlier post, I illustrated that any
Monoid instances can be unfolded, and wrote a generalized unfold function that works for any
You can see another example of the power this simple typeclass brings from this post. Now that you know what monoids are, you should be able to understand it.
I'm sure you will find more examples in your FP journey!
In the next post, we will deep dive another common typeclass:
Functor. You don't want to miss this one!
Software engineer. Loves FP Haskell Coq Agda PLT. Always learning. Prior: Economist. Vegan, WOC in solidarity with POC.
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