($) is the
function application operator.
Applying
($) 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