[Haskell-cafe] pointfree-trouble

[Felipe Lessa]

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 = (Matr .) ((flip (.) unMatr) a)
> reMatr a = (Matr .) $ (flip (.) unMatr) a
> reMatr = (Matr .) . (flip (.) unMatr)

--
Felipe.