Re: [Haskell-cafe] Rank N type tutorial?

Ben Rudiak-Gould wrote:

The way to think about it is that foralls are extra function arguments. Your first example is like

foo :: Integer -> (a::Type -> Show a -> a)

so a is chosen by the caller, not by you. The second case is like

bar :: Integer -> (a::Type -> Show a -> a -> b) -> b

In order for the first case to work as you expect, you'd need the type

foo :: Integer -> (a::Type, Show a, a)

which is traditionally written

foo :: Integer -> (exists a. Show a => a)

I thought "exists" was spelled "forall" in Haskell? Greg Buchholz