[Haskell-cafe] Rigid skolem type variable escaping scope
Erik Hesselink
hesselink at gmail.com
Wed Aug 22 22:32:04 CEST 2012
On Wed, Aug 22, 2012 at 10:13 PM, Matthew Steele <mdsteele at alum.mit.edu> wrote:
> On Aug 22, 2012, at 3:02 PM, Lauri Alanko wrote:
>
>> Quoting "Matthew Steele" <mdsteele at alum.mit.edu>:
>>
>>> {-# 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
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