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

[Svein Ove Aas]

Or if the specificity of (/\) is important to you, you could define f
at the global scope using that type, not export it, then declare (/\)
as equal to f but with a more restrictive type.

On Wed, Jan 21, 2009 at 5:45 AM, Alexander Dunlap
<alexander.dunlap at gmail.com> wrote:
> 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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>>
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