Haskell: Flipping a curried dollar operator -


say define function:

f = ($ 5)

then can apply it:

> f (\x -> x ^ 2) 25

its type is:

:t f f :: (integer -> b) -> b

which makes sense, gets function argument, , returns function applied on integer 5.

now define function:

g = flip f

i expect not make sense, because f function of single argument.

but, checking type:

:t g g :: b -> (integer -> b -> c) -> c

so g function of 2 arguments!

applying on values:

> g [2, 4, 6] (\x y -> x:y) [5,2,4,6]

what going on here? flip ($ 5) mean?

follow types:

($ 5) :: (int -> a) -> flip  :: (x -> y -> z) -> y -> x -> z

but since -> right associative, type x -> y -> z equivalent x -> (y -> z), so

flip  :: (x         -> (y -> z)) -> y -> x -> z ($ 5) :: (int -> a) ->

so x ~ (int -> a) , (y -> z) ~ a, substituting in:

($ 5) :: (int -> (y -> z)) -> (y -> z)

and simplified

($ 5) :: (int -> y -> z) -> y -> z

so

flip ($ 5) :: y -> (int -> y -> z) -> z

which equivalent type you're seeing (although used int instead of integer save typing).

what saying type of ($ 5) gets specialized when passed flip such takes function of 2 arguments. valid have ($ 5) const, const :: -> b -> a , ($ 5) const :: b -> int. ($ 5) doing applying 5 an argument function, not the argument function. example of partial application, not of arguments supplied function. that's why can things map (subtract 1) [1, 2, 3].

an example of how use flip ($ 5) is:

> flip ($ 5) 1 (**) 25.0 > flip ($ 5) 1 (-) 4.0 > let f x y = (x, y) > flip ($ 5) 1 f (5, 1)

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