I'm having trouble with understanding how function application works with currying in haskell. If I have following function:
($) :: (a -> b) -> a -> b
I understand that to partially apply this function I need to provide (a -> b) function ($'s first argument).
Why then is it possible to apply a value first (i.e. reverse arguments)?
($ 0) :: Num a => (a -> b) -> b
What am I missing here?
($) is an operator. In Haskell, any operator can be written in a left-section (like (x $)) or a right-section (like ($ x)):
(x $) = (\y -> x $ y) = ($) x
($ x) = (\y -> y $ x) = flip ($) x
Note that the only exception to this rule is (-), in order to conveniently write negative numbers:
\x -> (x-) :: Num a => a -> a -> a -- equivalent to \x -> (-) x
\x -> (-x) :: Num a => a -> a -- equivalent to \x -> negate x
In case you want to tersely write (\y -> y - x), you can use subtract:
\x -> subtract x :: Num a => a -> a -> a -- equivalent to \x -> flip (-) x