| Copyright | Nils Anders Danielsson 2006 Alexander Berntsen 2014 |
|---|---|
| License | BSD-style (see the LICENSE file in the distribution) |
| Maintainer | libraries@haskell.org |
| Stability | stable |
| Portability | portable |
| Safe Haskell | Trustworthy |
| Language | Haskell2010 |
Data.Function
Contents
- Prelude re-exports
- Other combinators
Description
Simple combinators working solely on and with functions.
Synopsis
- id :: a -> a
- const :: a -> b -> a
- (.) :: (b -> c) -> (a -> b) -> a -> c
- flip :: (a -> b -> c) -> b -> a -> c
- ($) :: forall (repa :: RuntimeRep) (repb :: RuntimeRep) (a :: TYPE repa) (b :: TYPE repb). (a -> b) -> a -> b
- (&) :: forall (r :: RuntimeRep) a (b :: TYPE r). a -> (a -> b) -> b
- fix :: (a -> a) -> a
- on :: (b -> b -> c) -> (a -> b) -> a -> a -> c
- applyWhen :: Bool -> (a -> a) -> a -> a
Prelude re-exports
const x y always evaluates to x, ignoring its second argument.
>>>const 42 "hello"42
>>>map (const 42) [0..3][42,42,42,42]
flip :: (a -> b -> c) -> b -> a -> c Source #
takes its (first) two arguments in the reverse order of flip ff.
>>>flip (++) "hello" "world""worldhello"
($) :: forall (repa :: RuntimeRep) (repb :: RuntimeRep) (a :: TYPE repa) (b :: TYPE repb). (a -> b) -> a -> b infixr 0 Source #
($) is the function application operator.
Applying ($) 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
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)
Uses
A common use cases of ($) is to avoid parentheses in complex expressions.
For example, instead of using nested parentheses in the following Haskell function:
-- | Sum numbers in a string: strSum "100 5 -7" == 98 strSum ::String->IntstrSum s =sum(mapMaybereadMaybe(wordss))
we can deploy the function application operator:
-- | Sum numbers in a string: strSum "100 5 -7" == 98 strSum ::String->IntstrSum s =sum$mapMaybereadMaybe$wordss
($) 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
Other combinators
(&) :: forall (r :: RuntimeRep) a (b :: TYPE r). a -> (a -> b) -> b infixl 1 Source #
is the least fixed point of the function fix ff,
i.e. the least defined x such that f x = x.
For example, 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,
>>>fix (\rec n -> if n <= 1 then 1 else n * rec (n-1)) 5120
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.
on :: (b -> b -> c) -> (a -> b) -> a -> a -> c infixl 0 Source #