[Haskell-cafe] Re: Looking for pointfree version (Kim-Ee Yeoh) (02/12)

[Gene Arthur]

>Kim-Ee Yeoh said:
>On the same note, does anyone have ideas 
>for the following snippet? Tried the pointfree 
>package but the output was useless. 

>pointwise op (x0,y0) (x1,y1) = (x0 `op` x1, y0 `op` y1)

First sorry for the delay in getting to this.. been behind on
projects so had some days of mail piled up.  Here is what 
I came up with, using one arrow operator,
so you would have to import: Control.Arrow ((***)) at minimum 
to use this solution:

pointfree :: 
    forall t t1 c. (t -> t1 -> c) -> (t, t1) -> (t, t1) -> (c, c)

pointfree op  = 
      curry $ (\(a,b) -> a `op` b)  *** (\(a,b) -> a `op` b)

examples of use, that were executed using:

ghci -fglasgow-exts -farrows Control.Arrow

*>pointfree (*) (3,5) (12,12)
(15,144)

*> pointfree (++) ("This ","That") ("Old"," Man")
("This That","Old Man")

Hope that helps.
 
-- gene
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