Re: [Haskell-cafe] Rigid skolem type variable escaping scope

On Wed, Aug 22, 2012 at 10:13 PM, Matthew Steele wrote:

On Aug 22, 2012, at 3:02 PM, Lauri Alanko wrote:

...

Quoting "Matthew Steele" :

...

{-# LANGUAGE Rank2Types #-}

class FooClass a where ...

foo :: (forall a. (FooClass a) => a -> Int) -> Bool foo fn = ...

...

newtype IntFn a = IntFn (a -> Int)

bar :: (forall a. (FooClass a) => IntFn a) -> Bool bar (IntFn fn) = foo fn

In case you hadn't yet discovered it, the solution here is to unpack the IntFn a bit later in a context where the required type argument is known:

bar ifn = foo (case ifn of IntFn fn -> fn)

Hope this helps.

Ah ha, thank you! Yes, this solves my problem.

However, I confess that I am still struggling to understand why unpacking earlier, as I originally tried, is invalid here. The two implementations are:

1) bar ifn = case ifn of IntFn fn -> foo fn 2) bar ifn = foo (case ifn of IntFn fn -> fn)

Why is (1) invalid while (2) is valid? Is is possible to make (1) valid by e.g. adding a type signature somewhere, or is there something fundamentally wrong with it? (I tried a few things that I thought might work, but had no luck.)

I can't help feeling like maybe I am missing some small but important piece from my mental model of how rank-2 types work. (-: Maybe there's some paper somewhere I need to read?

Look at it this way: the argument ifn has a type that says that *for any type a you choose* it is an IntFn. But when you have unpacked it by pattern matching, it only contains a function (a -> Int) for *one specific type a*. At that point, you've chosen your a. The function foo wants an argument that works for *any* type a. So passing it the function from IntFn isn't enough, since that only works for *one specific a*. So you pass it a case expression that produces a function for *any a*, by unpacking the IntFn only inside. I hope that makes sense (and is correct...) Erik