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 normal case (forall n. Num n => (n -> n) -> (Int, Double)), we choose an n first and then provide a function. So we could pass in a function of type Int -> Int, Double -> Double, Rational -> Rational and so on.

In the Rank 2 case ((forall n. Num n => n -> n) -> (Int, Double)) we have to provide the function before we know n. This means that the function has to work for any n; none of the examples I listed for the previous example would work.

We need this for the example code given because the function f that's passed in is applied to two different types: an Int and a Double. So it has to work for both of them.

The first case is normal because that's how type variables work by default. If you don't have a forall at all, your type signature is equivalent to having it at the very beginning. (This is called prenex form.) So Num n => (n -> n) -> (Int, Double) is implicitly the same as forall n. Num n => (n -> n) -> (Int, Double).

What's the type of a function that works for any n? It's exactly forall n. Num n => n -> n.

Answer 2

How do you "read" this former signature so that it sounds like what it means?

You can read

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

as "rankN takes a parameter f :: Num n => n -> n" and returns (Int, Double), where f :: Num n => n -> n can be read as "for any numeric type n, f can take an n and return an n".

The rank one definition

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

would then be read as "For any numeric type n, rank1 takes an argument f :: n -> n and returns an (Int, Double)".

Is the latter signature the same as simply Num n => (n -> n) -> (Int, Double) without the need for forall?

Yes, by default all foralls are implicitly placed at the outer-most position (resulting in a rank-1 type).

Answer 3

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