[Haskell-cafe] pointfree-trouble
Kim-Ee Yeoh
a.biurvOir4 at asuhan.com
Wed Dec 23 02:33:50 EST 2009
There you have it: fully- and semi-pointfree versions of reMatr.
A heads up: aggressively pursuing pointfreeness without type signatures
guarantees a courtesy call from the monomorphism restriction,
pace ()-garlic aficionados.
As for your question on why the original code doesn't typecheck: if
you explain how you arrived at it, perhaps we can figure out where
you tripped up. Daniel Fischer for instance, *calculated* for you
the right answer. Habeas calculus and all that.
slemi wrote:
>
> thanks, that's a really neat syntactic sugar :)
>
> however, my original question was how to make the reMatr function
> pointfree, as
> reMatr = Matr . (flip (.) unMatr)
> is not working. any ideas/explanation why it doesnt work?
>
>
> Kim-Ee Yeoh wrote:
>>
>> Here's another way of writing it:
>>
>> data Matrix a = Matr {unMatr :: [[a]]} | Scalar a deriving (Show, Eq)
>> -- RealFrac constraint removed
>>
>> reMatr :: RealFrac a => ([[a]] -> [[a]]) -> (Matrix a -> Matrix a)
>> reMatr f = Matr . f . unMatr -- this idiom occurs a lot, esp. with
>> newtypes
>>
>> Affixing constraints to type constructors is typically deprecated.
>>
>>
>>
>> slemi wrote:
>>>
>>> i have trouble making a function pointfree:
>>>
>>> data RealFrac a => Matrix a = Matr [[a]] | Scalar a
>>> deriving (Show, Eq)
>>>
>>> unMatr :: RealFrac a => Matrix a -> [[a]]
>>> unMatr = (\(Matr a) -> a)
>>>
>>> reMatr :: RealFrac a => ([[a]] -> [[a]]) -> (Matrix a -> Matrix a)
>>> reMatr a = Matr . (flip (.) unMatr) a
>>>
>>> this works fine, but if i leave the 'a' in the last function's
>>> definition like this:
>>> reMatr = Matr . (flip (.) unMatr)
>>> it gives an error. can anybody tell me why? (i'm using ghci)
>>>
>>
>>
>
>
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