[Haskell-cafe] Quite confused by simple transformations on this code not working

[Alexander Dunlap]

Instead of declaring (/\) :: Eq a => Sentence a -> Sentence a ->
Sentence a, you could say (/\) :: Eq a -> [a] -> [a] -> [a]. Then it
would work in both places. ([a] -> [a] -> [a] is a more general type
than [[Term a]] -> [[Term a]] -> [[Term a]], so functions with the
former type can be used in place of functions of the latter type but
not vice versa.)

Alex

2009/1/20 Andrew Wagner <wagner.andrew at gmail.com>:
> So...there's just no good way to avoid the duplication?
>
> On Tue, Jan 20, 2009 at 11:10 PM, wren ng thornton <wren at freegeek.org>
> wrote:
>>
>> Andrew Wagner wrote:
>>>
>>> Strange little bit of code:
>>> http://moonpatio.com:8080/fastcgi/hpaste.fcgi/view?id=829#a829
>>>
>>> If I do any of the following, all of which seem natural to me, it fails
>>> to
>>> typecheck:
>>>
>>>   1. move f out of the 'where' clause (with or without a type signature)
>>>   2. put the same type signature on f as is on (/\)
>>>   3. replace f with (/\) completely
>>>
>>> What's going on here?
>>
>>    > :t (nub .) . (++)
>>    (nub .) . (++) :: (Eq a) => [a] -> [a] -> [a]
>>
>>    > :t foldr (map . (nub .) . (++))
>>    foldr (map . (nub .) . (++)) :: (Eq a) => [[a]] -> [[a]] -> [[a]]
>>
>> The type you give to (/\) is more restrictive than the type of the
>> expression, and f uses the generality of the expression.
>>
>> --
>> Live well,
>> ~wren
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