Before we can get into the more advanced topics on Lenses, it is
important to really understand both `Foldable`

and `Traversable`

, which
is the motivation behind this article.

Let’s begin with `Foldable`

. `Foldable`

represents structures which can
be folded. What does that mean? Here are a few examples:

- Calculating the sum of a list.
- Calculating the product of a list.
- Folding a tree to get a maximum value.

We can describe a *fold* as ** taking a structure and reducing it to a
single result**. That’s also why some languages have a

`reduce`

function instead of a `fold`

, even though they mean the same thing. It is important to really understand the concept behind a fold in
general, not in terms of specific functions like `foldl`

or `foldr`

.
Whenever you see the word `fold`

in a function name, think *reducing a
larger structure to a single result*.

Now comes the time to take a look at the `Foldable`

type class.

```
class Foldable t where
fold :: Monoid m => t m -> m
foldMap :: Monoid m => (a -> m) -> t a -> m
foldr :: (a -> b -> b) -> b -> t a -> b
foldr' :: (a -> b -> b) -> b -> t a -> b
foldl :: (b -> a -> b) -> b -> t a -> b
foldl' :: (b -> a -> b) -> b -> t a -> b
foldr1 :: (a -> a -> a) -> t a -> a
foldl1 :: (a -> a -> a) -> t a -> a
```

We won’t go into detail on all of these, since `foldl`

, `foldr`

,
`foldl'`

, `foldr'`

, `foldl1`

and `foldr1`

work the same as their
counterparts from `Data.List`

.

What is interesting here is that `fold`

and `foldMap`

require the
elements of the `Foldable`

to be `Monoid`

s. Let’s just quickly take a
look at what a `Monoid`

is.

```
class Monoid a where
mempty :: a
mappend :: a -> a -> a
mconcat :: [a] -> a
```

Nothing really special here, `Monoid`

just simply defines a zero element
via `mempty`

and an associative operation `mappend`

for combining two
`Monoid`

s into one. `mconcat`

is just a convenience method which has a
default implementation using `mappend`

.

```
mconcat :: [a] -> a
mconcat = foldr mappend mempty
```

The interesting thing about `fold`

and `foldMap`

is that they use a
`Monoid`

instead of a function to give us the final result. This might
not be obvious at first, but by picking the right `Monoid`

it is
essentialy the same as passing in a function, since it will just use the
`mappend`

defined for that `Monoid`

instance.

One very very very important aspect to understand here is that it is the
`fold`

function that requires the elements of `Foldable`

to have a
`Monoid`

instance, while `Foldable`

itself does not have that
restriction.

The result of this is that we can have something like `[Int]`

, where the
`[]`

is a `Foldable`

, but `Int`

is not a `Monoid`

, though as long as we
don’t use any of the functions from `Foldable`

that require a `Monoid`

we’ll be OK. Here’s an example

```
λ> foldr1 (+) [1,2,3,4]
10
λ> fold ["hello", "world"]
"helloworld" -- Strings are Monoids using concatenation
λ> fold [1,2,3,4]
<interactive>:1:1:
No instance for (Monoid a0) arising from a use of ‘it’
```

See how the problem only arises when we used `fold`

with `Int`

. We could
however wrap those `Int`

s in a `Monoid`

such as `Sum`

or `Product`

and
`fold`

them then.

```
λ> fold [Sum 1, Sum 2, Sum 3, Sum 4]
Sum {getSum = 10}
```

This might seem tedious at first, but remember our `Foldable`

type
class, as it also defines a function that is perfect for this particular
use case: `foldMap :: Monoid m => (a -> m) -> t a -> m`

. We can read
this as *Given a foldable containing things that aren’t Monoids, and a
function that can convert a single thing to a Monoid, I’ll give you back
a Monoid by traversing the foldable, converting everything to Monoids
and folding them together.*

Here’s our previous example, but now using `foldMap`

.

```
λ> foldMap Sum [1,2,3,4]
Sum {getSum = 10}
```

If you think about this for a little while we might even implement
`fold`

in terms of `foldMap`

. Why? When using `foldMap`

we need to
provide a way to convert each item to a `Monoid`

, but if those items
already are `Monoid`

s, we don’t need to do any conversion!

```
fold :: Monoid m => t m -> m
fold xs = foldMap id xs
```

Here’s the same in more steps.

```
λ> :t id
:: a -> a
λ> :t fold
:: (Monoid m, Foldable t) => t m -> m
λ> :t foldMap
:: (Monoid m, Foldable t) => (a -> m) -> t a -> m
λ> :t foldMap id
:: (Monoid m, Foldable t) => t m -> m
```

The actual `Foldable`

type class requires either `foldMap`

or `foldr`

,
but for the sake of this article we won’t be looking into `foldr`

.

Now that we have an understanding of `Foldable`

we can move on to
something more fun, `Traversable`

. `Traversable`

represents data
structures which can be traversed while perserving the shape. This is
why there is no `filter`

or `concatMap`

, since `Traversable`

only
defines a way to move through the data structure, but not a way to
change it.

```
class (Functor t, Foldable t) => Traversable t where
traverse :: Applicative f => (a -> f b) -> t a -> f (t b)
sequenceA :: Applicative f => t (f a) -> f (t a)
```

If you look in the documentation for
`Traversable`

you might note that there is also `mapM`

and `sequence`

, but we won’t be
covering those in this article, since their implementation isn’t
interesting and can be done mechanically.

This might look a little intimidating at first, but don’t worry, we’ll
do this step by step by implementing a `Traversable`

instance for a
list.

```
instance Traversable [] where
traverse f xs = _
```

Since the implementation will be recursive we first need to define the
base case for our recursion, which will be the empty list. The type that
we’re looking for is `f [b]`

, but because the list we’re traversing is
empty, we just need to wrap it in the `Applicative`

context.

```
instance Traversable [] where
traverse _ [] = pure []
traverse f (x:xs) = _
```

Next goes the actual recursive implementaion. We have a function ```
f :: a
-> f b
```

and a head of the list which has the type `a`

. The only thing we
can do at this point is apply the function.

```
instance Traversable [] where
traverse _ [] = pure []
traverse f (x:xs) = f x
```

This won’t typecheck of course, because we’re returning `f b`

instead of
`f [b]`

. We could cheat here a little bit and just try to apply a some
function `f b -> f [b]`

to get the result. We can use `(:[])`

which has
a type of `a -> [a]`

and `fmap`

it on what we have.

```
instance Traversable [] where
traverse _ [] = pure []
traverse f (x:xs) = fmap (:[]) (f x)
```

Now we have an implementation that type checks, but it is still wrong,
since it doesn’t satisfy the rule that *a traversal must not change the
shape of the structure it is traversing*, and here we are just dropping
the rest of the list. We need to find a way to use recursion and somehow
combine the results.

By looking at the type of `traverse :: (a -> f b) -> t a -> f (t b)`

, or
in our case specifically `traverse :: (a -> f b) -> [a] -> f [b]`

we can
see that using `traverse`

recursively on the tail of the list would give
us the type we need.

```
instance Traversable [] where
traverse _ [] = pure []
traverse f (x:xs) = (f x) _ traverse f xs
```

Now we have two values, one of type `f b`

and one of type `f [b]`

, which
are basically the head and the tail of the list, both wrapped in an
`Applicative`

context. We also have a function `(:) :: a -> [a] -> [a]`

,
which concatenates a head and a tail together into a single list.

Knowing all of this it just comes down to a basic use of `Applicative`

where we have a function of two arguments and need to apply it to two
values in the `Applicative`

context. We can do this in two different
ways.

```
instance Traversable [] where
traverse _ [] = pure []
traverse f (x:xs) = (:) <$> f x <*> traverse f xs
```

And an alternative definition using `liftA2`

.

```
instance Traversable [] where
traverse _ [] = pure []
traverse f (x:xs) = liftA2 (:) (f x) (traverse f xs)
```

It should be pretty clear now that we need the `Applicative`

to be able
to actually implement `traverse`

. If all we had was a `Functor`

we
wouldn’t be able to combine the `f b`

and `f [b]`

together.

Now that we have `traverse`

we can move on to define `sequenceA`

. Here’s
a specific type for our list instance.

```
sequenceA :: Applicative f => [f a] -> f [a]
```

If you’re familiar with `sequence :: Monad m => [m a] -> m [a]`

from
`Control.Monad`

then you can see how these two functions are doing the
same thing. It simply takes the `Applicative`

effects, runs them and
pulls them out of the list.

The implementation is really simple. Starting out with an empty list, we
just need to wrap it in the `Applicative`

context.

```
sequenceA [] = pure []
```

Next comes the actual recursive implementaiton. If we pattern match on
the head and the tail of the list, we’ll yet again get `f a`

and ```
[f
a]
```

.

```
sequenceA (x:xs) = _
```

We can call `sequenceA`

recursively on the tail to get `f [a]`

.

```
sequenceA [] = pure []
sequenceA (x:xs) = sequenceA xs
```

But of course this isn’t good enough. We need a way to combine the head
and the tail while they’re both wrapped in an `Applicative`

context.
This can be done in the same way as we did previously with `traverse`

,
using `(:)`

and the `Applicative`

functions `<$>`

and `<*>`

.

```
sequenceA [] = pure []
sequenceA (x:xs) = (:) <$> x <*> sequenceA xs
```

Or alternatively using `liftA2`

again.

```
sequenceA [] = pure []
sequenceA (x:xs) = liftA2 (:) x (sequenceA xs)
```

That’s it, we have a working implementation for `sequenceA`

.

If we now look at our implementations for `traverse`

and `sequenceA`

we
can definitely see some similarity there.

```
traverse _ [] = pure []
traverse f (x:xs) = (:) <$> f x <*> traverse f xs
sequenceA [] = pure []
sequenceA (x:xs) = (:) <$> x <*> sequenceA xs
```

The only difference is that `traverse`

takes a function and applies it
to the head of the list, while `sequenceA`

simply uses the head as it
is. Knowing this we can actually define `sequenceA`

using `traverse`

and
the `id`

function.

```
sequenceA :: (Traversable t, Applicative f) => t (f a) -> f (t a)
sequenceA xs = traverse id xs
```

Could we do the same thing the other way around though? Yes! We most
certainly can define `traverse`

by using `sequenceA`

and the fact that
every `Traversable`

is also a `Functor`

. Let’s take this step by step.

```
traverse :: (Traversable t, Applicative f) => (a -> f b) -> t a -> f (t b)
traverse f xs = _
```

We only have one way of applying our function `a -> f b`

to the `t a`

and that is using `fmap`

, which would give us `t (f b)`

.

```
traverse :: (Traversable t, Applicative f) => (a -> f b) -> t a -> f (t b)
traverse f xs = _ $ fmap f xs
```

Now we’ll get an error saying that we need a function ```
t (f b) -> f (t
b)
```

, which is exactly what `sequenceA`

does!

```
traverse :: (Traversable t, Applicative f) => (a -> f b) -> t a -> f (t b)
traverse f xs = sequenceA $ fmap f xs
```

Given the two implementations we just got we can rewrite our initial
`Traversable`

type class to use those as a default implementation for
both functions.

```
class (Functor t, Foldable t) => Traversable t where
traverse :: Applicative f => (a -> f b) -> t a -> f (t b)
traverse f xs = sequenceA $ fmap f xs
sequenceA :: Applicative f => t (f a) -> f (t a)
sequenceA xs = traverse id xs
```

This is actually how it’s done in the `Data.Traversable`

module, except
that if you look at the source code you’ll see the functions defined in
point free style.

```
class (Functor t, Foldable t) => Traversable t where
traverse :: Applicative f => (a -> f b) -> t a -> f (t b)
traverse f = sequenceA . fmap f
sequenceA :: Applicative f => t (f a) -> f (t a)
sequenceA = traverse id
```

It might not be so obvious at first, but a `Traversable`

is a very
powerful concept. So powerful that it actually allows us to define both
`Functor`

and `Foldable`

if we have just a single function from
`Traversable`

. The `Data.Traversable`

module defines two functions,
`fmapDefault`

and `foldMapDefault`

, which can be used as an
implementation for `fmap`

and `foldMap`

if we so desire.

```
fmapDefault :: Traversable t => (a -> b) -> t a -> t b
foldMapDefault :: (Traversable t, Monoid m) => (a -> m) -> t a -> m
```

The way we’re going to implement these is very similar to what we did in the Lens introduction article. If this section is too hard for you to understand I recommend reading the Lens article first and then come back here. Everything will make a lot more sense.

Let’s first compare the types of `traverse`

and `fmap`

.

```
fmap :: Functor f => (a -> b) -> f a -> f b
traverse :: (Traversable t, Applicative f) => (a -> f b) -> t a -> f (t b)
```

The difference is that the function passed to `traverse`

returns a value
wrapped in `Applicative`

context, and the result is also wrapped. If we
could find a way to wrap the value after we apply the function, and then
unwrap it at the end, we would get exactly the same type as `fmap`

.

We can use the `Identity`

functor to do this, which defines a way to
unwrap it using `runIdentity :: Identity a -> a`

.

```
fmapDefault :: Traversable t => (a -> b) -> t a -> t b
fmapDefault f x = _
```

We don’t have that many options here. To be able to give the function
`f`

to a `traverse`

we need to change it’s type from `a -> f a`

. That’s
where `Identity`

comes in.

```
fmapDefault :: Traversable t => (a -> b) -> t a -> t b
fmapDefault f x = traverse (Identity . f) x
```

Now our types don’t align, since we are supposed to return `t b`

but we
are returning `Identity (t b)`

. The solution here is the above mentioned
`runIdentity`

which simply unwraps the value.

```
fmapDefault :: Traversable t => (a -> b) -> t a -> t b
fmapDefault f x = runIdentity $ traverse (Identity . f) x
```

And once more in point free style.

```
fmapDefault :: Traversable t => (a -> b) -> t a -> t b
fmapDefault f = runIdentity . traverse (Identity . f)
```

Compare this to the definition of `over`

and you can see how it *looks
and feels* almost exactly the same.

```
over :: Lens s a -> (a -> a) -> s -> s
over ln f = runIdentity . ln (Identity . f)
```

We’ll explain how this relates to Lenses in more detail in a followup
article, but for now let’s move on to `foldMapDefault`

.

This part is very hard to understand, so be careful.

If we compare the type of `foldMapDefault`

with `traverse`

we can yet
again see some similarity.

```
foldMapDefault :: (Traversable t, Monoid m) => (a -> m) -> t a -> m
traverse :: (Traversable t, Applicative f) => (a -> f b) -> t a -> f (t b)
```

The difference from `fmapDefault`

is that now we need a way to convert
each element of the `Traversable`

to a `Monoid`

.

We will use the `Const`

applicative here, which as it so happens also
defines a `Monoid`

instance.

```
foldMapDefault :: (Traversable t, Monoid m) => (a -> m) -> t a -> m
foldMapDefault f x = _
```

As previously we can only use `traverse`

together with a function ```
a ->
f b
```

, but we have `a -> m`

, where by using `Const`

we can do the ```
m -> f
b
```

.

```
foldMapDefault :: (Traversable t, Monoid m) => (a -> m) -> t a -> m
foldMapDefault f x = traverse (Const . f) x
```

Again we’re faced with the problem of having `Const m (t b)`

instead of
`m`

, which can be solved using `getConst`

.

```
foldMapDefault :: (Traversable t, Monoid m) => (a -> m) -> t a -> m
foldMapDefault f x = getConst $ traverse (Const . f) x
```

And a point free version.

```
foldMapDefault :: (Traversable t, Monoid m) => (a -> m) -> t a -> m
foldMapDefault f = getConst . traverse (Const . f)
```

This is also very similar to one of the functions Lens provides, in
particular `view`

.

```
view :: Lens s a -> s -> a
view ln = getConst . ln Const
```

Now that we understand how both `fmapDefault`

and `foldMapDefault`

work,
we can use them to define a `Functor`

and a `Foldable`

instance for any
`Traversable`

we might have.

We can test this out by defining a simple list type.

```
data List a = Nil
| Cons a (List a)
deriving Show
instance Functor List where
fmap = fmapDefault
instance Foldable List where
foldMap = foldMapDefault
instance Traversable List where
traverse _ Nil = pure Nil
traverse f (Cons x xs) = fmap Cons (f x) <*> traverse f xs
```

We used `fmap = fmapDefault`

and `foldMap = foldMapDefault`

to define
our `Functor`

and `Foldable`

instances, which is all made possible by
also having a `Traversable`

instance. Let’s test this out to make sure
it works!

```
λ> traverse (\x -> Just (x + 1)) (Cons 1 (Cons 2 (Cons 3 Nil)))
Just (Cons 2 (Cons 3 (Cons 4 Nil)))
λ> fold (Cons "hello" (Cons "world" Nil))
"helloworld"
λ> fmap (+1) (Cons 1 (Cons 2 (Cons 3 Nil)))
Cons 2 (Cons 3 (Cons 4 Nil))
```

It might be a surprising, but everything works as it is supposed to.

Subscribe to receive updates and free content from the book. You'll also get a discount when the final version of the book is released.

- Fibonacci Numbers
- Parsing CSS with Parsec
- Lens Tutorial - Stab & Traversal (Part 2)
- Foldable and Traversable
- Building Monad Transformers - Part 1
- Mutable State in Haskell
- Lens Tutorial - Introduction (part 1)
- Using Phantom Types in Haskell for Extra Safety - Part 2
- Using Phantom Types for Extra Safety
- Evil Mode: How I Switched From VIM to Emacs