forall a (Ord a => a-> a) -> Int is an illegal type???
Stefan Holdermans
stefan at cs.uu.nl
Fri Feb 10 10:07:08 EST 2006
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
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