forall a (Ord a => a-> a) -> Int is an illegal type???

[Stefan Holdermans]

Brian,


>>> Also, the rule would not be quite as simple as you make it out to  
>>> be,
>>> since
>>>
>>>      forall a. (forall b. Foo a b => a -> b) -> Int
>>>
>>> is a legal type, for example.
>>
>> Is it? GHCi gives me an error if I try typing a function like that.
>>
>>> {-# OPTIONS -fglasgow-exts #-}
>>> class Foo a b
>>>
>>> f :: forall a. (forall b. Foo a b => a -> b) -> Int
>>> f = undefined
>>
>>     No instance for (Foo a b)
>>       arising from instantiating a type signature at x.hs:5:4-12
>>     Probable fix: add (Foo a b) to the type signature(s) for `f'
>>       Expected type: (forall b1. (Foo a b1) => a -> b1) -> Int
>>       Inferred type: (a -> b) -> Int
>>     In the definition of `f': f = undefined

Short answer: forall a. a -> a cannot be instantiated to the type of  
f. Try:

   {-# OPTIONS -fglasgow-exts #-}
   class Foo a b
   f   :: forall a. (forall b. Foo a b => a -> b) -> Int
   f _ =  undefined


Now, you should be fine.

HTH,

   Stefan