Haskell's ($) is a magic operator?

Question

Say I have the following functions:

infixr 0 <|

{-# INLINE (<|) #-}
(<|) :: (a -> b) -> a -> b
f <| x = f x

foo :: a -> (forall b.          


        

Answer 1

Yes, there's a small amount of compiler magic around ($) to handle impredicative types. It was introduced because everyone expects

runST $ do
  foo
  bar
  baz

to typecheck, but it cannot normally. For more details see here (search runST), this email, and this email. The short of it is that there's actually a special rule in the type-checker specifically for ($) which gives it the ability to resolve the common case of impredicative types.