This example seems to break the type sig for $, and it works [duplicate]

Question

If the type for ($) is (a -> b) -> a -> b, then why are you allowed to curry it as ($2)? 2 is not of type (a -> b). See below example.

map ($2)[(+1),(+2)]

This is legit, awesome and intuitively makes sense. Please tell me how it is consistent with the type system rules?

Cheers

Answer 1

The behavior you are observing is due to how partial application works for infix operators. This is often called "section application" and you are applying 2 as the "right section" which would be the second argument. So you have:

($) :: (a -> b) -> a -> b
                   ^
                   |
                  This is the type variable for the argument '2'

And you can confirm this via:

ghci
> :t ($2)
($2) :: Num a => (a -> b) -> b

You can likely find this info hidden somewhere in most decently complete tutorials or you can see the Haskell report section on sections.