Understanding Haskell's RankNTypes

Question

While working my way through GHC extensions, I came across RankNTypes at the School of Haskell, which had the following example:

main = print $ rankN (+1)

r         


        

Answer 1

In the rankN case f has to be a polymorphic function which is valid for all numeric types n.

In the rank1 case f only has to be defined for a single numeric type.

Here is some code which illustrates this:

{-# LANGUAGE RankNTypes #-}

rankN :: (forall n. Num n => n -> n) -> (Int, Double)
rankN = undefined

rank1 :: forall n. Num n => (n -> n) -> (Int, Double)
rank1 = undefined

foo :: Int -> Int  -- monomorphic
foo n = n + 1

test1 = rank1 foo -- OK

test2 = rankN foo -- does not type check

test3 = rankN (+1) -- OK since (+1) is polymorphic

Update

In response to @helpwithhaskell's question in the comments...

Consider this function:

bar :: (forall n. Num n => n -> n) -> (Int, Double) -> (Int, Double)
bar f (i,d) = (f i, f d)

That is, we apply f to both an Int and a Double. Without using RankNTypes it won't type check:

-- doesn't work
bar' :: ??? -> (Int, Double) -> (Int, Double)
bar' f (i,d) = (f i, f d)

None of the following signatures work for ???:

Num n => (n -> n)
Int -> Int
Double -> Double