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

On Aug 22, 2012, at 4:32 PM, Erik Hesselink wrote:

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

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

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Quoting "Matthew Steele" :

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{-# 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.

Okay, that does make sense, and I can see now why (2) works while (1) doesn't. So my next question is: why does unpacking the newtype via pattern matching suddenly limit it to a single monomorphic type? As you said, ifn has the type "something that can be an (IntFn a) for any a you choose". Since an (IntFn a) is just a newtype around an (a -> Int), why, when you unpack ifn into (IntFn fn), is the type of fn not "something that can be an (a -> Int) for any a you choose"? Is it possible to add a local type annotation to force fn to remain polymorphic? Cheers, -Matt