# Functors

**Functors**

*generalise the idea of*

**function mapping**on a data structure type, i.e. applying a function to each element in the data structure. We have looked at the

`map`

function that does exactly that but is limited to a specific data structure - `list`

. However, any **parameterised**

**type**

*(data structure) can be made into an instance of the*

`functor`

class so that it supports function mapping to its elements through the use of the `fmap`

function:class Functor f where

fmap :: (a -> b) -> f a -> f b

(<$) :: a -> f b -> f a

-- Minimal complete definition:

-- fmap

The

`(<$)`

function simply replaces all elements in the data structure with the given value `a.`

Comparing

`fmap`

to `map`

:map :: (a -> b) -> [a] -> [b]

We can see that

`fmap`

has the exact same type signature as `map`

- but`map`

is limited to the type of `lists`

, whereas `fmap`

can be used with any parameterised type that uses a *type constructor*`f`

, i.e. is a class of `Functor f`

. Note that the form of`[a]`

and `[b]`

is just syntax sugar for the list type and is equivalent to `[] a`

and `[] b`

, where `[]`

is the *type constructor*for the list type.Since the

`map`

and `fmap`

functions perform the same action, it is very simple to make the `list`

type into a functor (and it is made into one by default) by using the `instance` keyword:instance Functor [] where

-- fmap :: (a -> b) -> [] a -> [] b

fmap = map

However, when it comes to other data structure types, we have to define the

`fmap`

function ourselves. For example, the `Maybe`

type (also an instance of the `Functor`

class by default) can be made an instance of the `Functor`

class as:instance Functor Maybe where

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

fmap _ Nothing = Nothing

fmap f (Just x) = Just (f x)

In the case of a

`Nothing`

, we ignore the function and return `Nothing`

as `Nothing`

generally represents an error state that we want to propagate through our `fmap`

function as well. Otherwise, we apply the function `f`

to the underlying value `x`

of `Just x`

and return the `Maybe`

type (using the `Just`

constructor) of the result, i.e. the function application `f x`

.Now let's see how we can make a newly-defined data type into an instance of the

`Functor`

class. We will first define a data type `Tree`

that will represent a full **binary tree**. A full binary tree is a binary tree type in which every tree node has either`0`

or `2`

children. In our case, a node with `0`

children will be a `Leaf`

that stores some value, and a node with `2`

children will be a `Branch`

that branches into two new nodes without storing a value itself:data Tree a = Leaf a | Branch (Tree a) (Tree a)

deriving (Show)

And now, we want to make

`Tree`

a functor, so we have to define the `fmap`

function for it:instance Functor Tree where

-- fmap :: (a -> b) -> Tree a -> Tree b

fmap f (Leaf x) = Leaf (f x)

fmap f (Branch l r) = Branch (fmap f l) (fmap f r)

The first case of

`Leaf`

is analogous to how we defined the `Just`

case of the `Maybe`

functor - we apply the function `f`

to the underlying value `x`

and return it wrapped in the `Leaf`

constructor. In the other case, where we are dealing with a `Branch`

we recursively apply `fmap`

to both children nodes of the branch.To represent the following binary tree:

Example binary tree.

We can use the following code and

`fmap`

over the entire data structure, applying the specified function (in this case `(/2)`

) to each leaf:ghci> myTree = (Branch (Branch (Leaf 1) (Leaf 2)) (Branch (Leaf 10) (Leaf 20)))

ghci> fmap (/2) myTree

Branch (Branch (Leaf 0.5) (Leaf 1.0)) (Branch (Leaf 5.0) (Leaf 10.0))

Note that the

`Prelude`

comes with an infix operator for `fmap`

as well - `(<$>)`

, not to be confused with the function application operator, `($)`

: ($) :: (a -> b) -> a -> b

(<$>) :: Functor f => (a -> b) -> f a -> f b

(<$>) = fmap

`($)`

is the simple function application operator we learned about before, while `(<$>)`

is *also a function application operator*but with lifting over a Functor. We could also write the above code as:

ghci> myTree = (Branch (Branch (Leaf 1) (Leaf 2)) (Branch (Leaf 10) (Leaf 20)))

ghci> (/2) <$> myTree

Branch (Branch (Leaf 0.5) (Leaf 1.0)) (Branch (Leaf 5.0) (Leaf 10.0))

There are two functor laws that must be satisfied and ensure that

`fmap`

works as intended:fmap id = id

The

`id`

is the identity function, and it is used to simply return the unaltered argument passed in. This means that mapping the `id`

function over a structure should return the same structure back unaltered.fmap (g . f) = fmap g . fmap f

The second functor law ensures that

*function**composition*is preserved when using`fmap`

so that it does not matter whether we map a composed function `(g . f)`

or map the first function `g`

and then the second function `f`

, as long as the order of the `g`

and `f`

functions stays the same.