The type of the Prelude function flip is:
flip :: (a -> b -> c) -> b -> a -> c
I.e., it takes one binary function and two arguments.
The type of the Prelude function id is:
id :: a -> a
But the type of flip id is:
flip id :: a -> (a -> b) -> b
How is it possible to apply flip to id when id is a unary function and flip requires binary function for the first arg?
btw. flip id is similar to x f -> f x
Haskell makes id fit the type of the first argument to flip by setting a = b -> c. So:
flip :: ( a -> b -> c) -> b -> a -> c
flip :: ((b -> c) -> b -> c) -> b -> (b -> c) -> c
flip id :: b -> (b -> c) -> c
where id is taken to be of type
id :: (b -> c) -> b -> c
which is equivalent to
id :: (b -> c) -> (b -> c)
i.e. a specialisation of id that only applies to unary functions.
Edit: I think I might rephrase my first line as:
Haskell deduces that id fits the type of the first argument to flip if a = b -> c.
In case that s any clearer.
Nefrubyr explains it very well.
Another way to (hopefully) make this a bit more intuitive is to think of the function application operator ($).
($) is a specialized form of id:
($) :: (a -> b) -> (a -> b)
($) = id
I ve seen the definition (#) = flip ($), such that you can write the argument before the function its applied to: obj # show.
Obviously, since ($) is just a specialized form of id, you could also write: (#) = flip id
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