[Haskell-cafe] pointfree-trouble

[Kim-Ee Yeoh]

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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