What is "n" in RankNTypes

foo has one argument that includes a universal quantor, that what kicks in the need for RankN. But this argument's type itself, a -> a, is rank-1, it's the only argument, so foo has rank n with n − 1 = 1, i.e. foo is rank-2.

Now consider

bar :: ((forall a. a -> a) -> (Char,Bool)) -> Int

This has an argument of foo's type, which as we said has Rank 2. So that's the highest rank in bar's arguments; bar is thus a rank-3 function.

n is the level at which the forall(s) is/are nested. So if you have forall a. ((a -> a) -> Bla) (which is simply a more verbose way of writing (a -> a) -> Bla), then the forall is on the outside and applies to the whole function, so it's rank 1. With (forall a. a -> a) -> Bla the forall only applies to the inner function (i.e. the one you take as an argument) and is thus rank 2.

If the function that you take as an argument would itself take another function as an argument and that function would have a forall in its type, then that would be rank 3. And so on.

Rank is defined inductively on the structure of types:

rank (forall a. T) = max 1 (rank T)
rank (T -> U)      = max (if rank T = 0 then 0 else rank T + 1) (rank U)
rank (a)           = 0

Note how it increases by one on the left-hand side of an arrow. So:

Rank 0: Int
Rank 1: forall a. a -> Int
Rank 2: (forall a. a -> Int) -> Int
Rank 3: ((forall a. a -> Int) -> Int) -> Int

and so on.