Data.Function

Identity function.

id x = x

This function might seem useless at first glance, but it can be very useful in a higher order context.

Examples

>>> length $ filter id [True, True, False, True]3

>>> Just (Just 3) >>= idJust 3

>>> foldr id 0 [(^3), (*5), (+2)]1000

const x y always evaluates to x, ignoring its second argument.

const x = \_ -> x

This function might seem useless at first glance, but it can be very useful in a higher order context.

Examples

>>> const 42 "hello"42

>>> map (const 42) [0..3][42,42,42,42]

Right to left function composition.

prop> (f . g) x = f (g x)

prop> f . id = f = id . f

Examples

>>> map ((*2) . length) [[], [0, 1, 2], [0]][0,6,2]

>>> foldr (.) id [(+1), (*3), (^3)] 225

>>> let (...) = (.).(.) in ((*2)...(+)) 5 1030

flip f takes its (first) two arguments in the reverse order of f.

prop> flip f x y = f y x

prop> flip . flip = id

Examples

>>> flip (++) "hello" "world""worldhello"

>>> let (.>) = flip (.) in (+1) .> show $ 5"6"

($) is the function application operator.

Applying ($) to a function f and an argument x gives the same result as applying f to x directly. The definition is akin to this:

($) :: (a -> b) -> a -> b
($) f x = f x

This is id specialized from a -> a to (a -> b) -> (a -> b) which by the associativity of (->) is the same as (a -> b) -> a -> b.

On the face of it, this may appear pointless! But it's actually one of the most useful and important operators in Haskell.

The order of operations is very different between ($) and normal function application. Normal function application has precedence 10 - higher than any operator - and associates to the left. So these two definitions are equivalent:

expr = min 5 1 + 5
expr = ((min 5) 1) + 5

($) has precedence 0 (the lowest) and associates to the right, so these are equivalent:

expr = min 5 $ 1 + 5
expr = (min 5) (1 + 5)

Examples

A common use cases of ($) is to avoid parentheses in complex expressions.

For example, instead of using nested parentheses in the following Haskell function:

-- | Sum numbers in a string: strSum "100  5 -7" == 98
strSum :: String -> Int
strSum s = sum (mapMaybe readMaybe (words s))

we can deploy the function application operator:

-- | Sum numbers in a string: strSum "100  5 -7" == 98
strSum :: String -> Int
strSum s = sum $ mapMaybe readMaybe $ words s

($) is also used as a section (a partially applied operator), in order to indicate that we wish to apply some yet-unspecified function to a given value. For example, to apply the argument 5 to a list of functions:

applyFive :: [Int]
applyFive = map ($ 5) [(+1), (2^)]
>>> [6, 32]

Technical Remark (Representation Polymorphism)

($) is fully representation-polymorphic. This allows it to also be used with arguments of unlifted and even unboxed kinds, such as unboxed integers:

fastMod :: Int -> Int -> Int
fastMod (I# x) (I# m) = I# $ remInt# x m

& is a reverse application operator. This provides notational convenience. Its precedence is one higher than that of the forward application operator $, which allows & to be nested in $.

This is a version of flip id, where id is specialized from a -> a to (a -> b) -> (a -> b) which by the associativity of (->) is (a -> b) -> a -> b. flipping this yields a -> (a -> b) -> b which is the type signature of &

Examples

>>> 5 & (+1) & show"6"

>>> sqrt $ [1 / n^2 | n <- [1..1000]] & sum & (*6)3.1406380562059946

fix f is the least fixed point of the function f, i.e. the least defined x such that f x = x.

When f is strict, this means that because, by the definition of strictness, f ⊥ = ⊥ and such the least defined fixed point of any strict function is ⊥.

Examples

We can write the factorial function using direct recursion as

>>> let fac n = if n <= 1 then 1 else n * fac (n-1) in fac 5120

This uses the fact that Haskell’s let introduces recursive bindings. We can rewrite this definition using fix,

Instead of making a recursive call, we introduce a dummy parameter rec; when used within fix, this parameter then refers to fix’s argument, hence the recursion is reintroduced.

>>> fix (\rec n -> if n <= 1 then 1 else n * rec (n-1)) 5120

Using fix, we can implement versions of repeat as fix . (:) and cycle as fix . (++)

>>> take 10 $ fix (0:)[0,0,0,0,0,0,0,0,0,0]

>>> map (fix (\rec n -> if n < 2 then n else rec (n - 1) + rec (n - 2))) [1..10][1,1,2,3,5,8,13,21,34,55]

Implementation Details

The current implementation of fix uses structural sharing

fix f = let x = f x in x

A more straightforward but non-sharing version would look like

fix f = f (fix f)

on b u x y runs the binary function b on the results of applying unary function u to two arguments x and y. From the opposite perspective, it transforms two inputs and combines the outputs.

(op `on` f) x y = f x `op` f y

Examples

>>> sortBy (compare `on` length) [[0, 1, 2], [0, 1], [], [0]][[],[0],[0,1],[0,1,2]]

>>> ((+) `on` length) [1, 2, 3] [-1]4

>>> ((,) `on` (*2)) 2 3(4,6)

Algebraic properties

• (*) `on` id = (*) -- (if (*) ∉ {⊥, const ⊥})

• ((*) `on` f) `on` g = (*) `on` (f . g)

• flip on f . flip on g = flip on (g . f)

applyWhen applies a function to a value if a condition is true, otherwise, it returns the value unchanged.

It is equivalent to flip (bool id).

Examples

>>> map (\x -> applyWhen (odd x) (*2) x) [1..10][2,2,6,4,10,6,14,8,18,10]

>>> map (\x -> applyWhen (length x > 6) ((++ "...") . take 3) x) ["Hi!", "This is amazing", "Hope you're doing well today!", ":D"]["Hi!","Thi...","Hop...",":D"]

Algebraic properties

• applyWhen True = id

• applyWhen False f = id