Am Freitag 25 Dezember 2009 00:19:58 schrieb Mark Lentczner:
> For the record, all of these compile with -O to exactly the same code!
>
> > reMatr1 f m = Matr (f (unMatr m))
> > reMatr2 f m = Matr $ f $ unMatr m
> > reMatr3 f m = Matr . f . unMatr $ m
> > reMatr4 f = Matr . f . unMatr
> > reMatr5
For the record, all of these compile with -O to exactly the same code!
> reMatr1 f m = Matr (f (unMatr m))
> reMatr2 f m = Matr $ f $ unMatr m
> reMatr3 f m = Matr . f . unMatr $ m
> reMatr4 f = Matr . f . unMatr
> reMatr5 f = Matr . (flip (.) unMatr) f
> reMatr6 = (Matr .) . (. unMatr)
A
Am Mittwoch 23 Dezember 2009 14:40:46 schrieb slemi:
> i dont know any calculus-thingy, this is what i did:
>
> reMatr a = Matr . a . unMatr
> reMatr a = Matr . (. unMatr) a
> reMatr a = Matr . (flip (.) unMatr) a
You need to be aware of the implicit parentheses, that is
Matr . ((flip (.) unMatr)
i dont know any calculus-thingy, this is what i did:
reMatr a = Matr . a . unMatr
reMatr a = Matr . (. unMatr) a
reMatr a = Matr . (flip (.) unMatr) a
reMatr = Matr . (flip (.) unMatr)
as http://old.nabble.com/pointfree-trouble-td26881661.html#a26889388 Daniel
pointed out, this doesnt work bec
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
y
Am Dienstag 22 Dezember 2009 15:09:34 schrieb slemi:
> hello everybody, i'm a newbie this is my first post here..
>
> 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 =
On Tue, Dec 22, 2009 at 3:09 PM, slemi <0sle...@gmail.com> wrote:
> this works fine, but if i leave the 'a' in the last function's definition
> like this:
> reMatr = Matr . (flip (.) unMatr)
The correct point free version would be :
> reMatr = (Matr .) . (. unMatr)
--
Jedaï
On Tue, Dec 22, 2009 at 09:27:47AM -0500, Keith Sheppard wrote:
> Hello, I didn't try to understand what the function is doing, but just
> quickly noticed that
>
> > reMatr a = Matr . (flip (.) unMatr) a
>
> can be written as
> > reMatr a = Matr . ((flip (.) unMatr) a)
...and then
> reMatr a = (M
On Tue, Dec 22, 2009 at 12:50:26AM -0800, Kim-Ee Yeoh wrote:
> reMatr :: RealFrac a => ([[a]] -> [[a]]) -> (Matrix a -> Matrix a)
> reMatr f = Matr . f . unMatr -- this idiom occurs a lot, esp. with
> newtypes
And usually we would call this 'liftMatr' or something along
these lines. The func
Hello, I didn't try to understand what the function is doing, but just
quickly noticed that
> reMatr a = Matr . (flip (.) unMatr) a
can be written as
> reMatr a = Matr . ((flip (.) unMatr) a)
but that
> reMatr = Matr . (flip (.) unMatr)
can be written as
> reMatr a = (Matr . (flip (.) unMatr))
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
hello everybody, i'm a newbie this is my first post here..
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]]) -> (Matr
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 con
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