On Aug 22, 2012, at 3:02 PM, Lauri Alanko wrote:
> Quoting "Matthew Steele" <mdste...@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?
Cheers,
-Matt
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