[Haskell-cafe] Head and tail matching question

[Simon Brenner]

Why not do something like this instead?

untab [] = []
untab xs = head : untab (drop 1 tail)
    where (head, tail) = break (== '\t') xs

BTW, going the extra step through unfoldr seems unnecessary to me - is
there any special reason to prefer unfolds over simple recursive
functions here? (Of course, you do get rid of the explicit recursive
call to untab - but in turn you have to run it all through unfoldr..)

Another, more pointless way is:

untab [] = []
untab xs = uncurry (:) $ second (untab . drop 1) $ break (== '\t') xs

(needs the additional import of Control.Arrow to get second)

On 6/11/07, Jules Bean <jules at jellybean.co.uk> wrote:
> Olivier Boudry wrote:
> > Hi all,
> >
> > I'm trying to write a untab function that would split a string on tabs
> > and return a list. Code is here.
> >
> > import Data.List (break, unfoldr)
> > import Data.Char (String)
> >
> > untab :: String -> [String]
> > untab s = unfoldr untab' s
> >
> > untab' :: String -> Maybe (String, String)
> > untab' s | s == "" = Nothing
> >         | otherwise = Just (h, ts)
> >         where (h, t:ts) = break (== '\t') s
> >
> > This code raises an exception when handling the last portion of the
> > string. Break returns a ("something", "") and t:ts cannot match on "".
>
>
> untab' [] = Nothing
> untab' s  = Just (h , drop 1 t)
>    where (h,t) = break (== '\t') s
> _______________________________________________
> Haskell-Cafe mailing list
> Haskell-Cafe at haskell.org
> http://www.haskell.org/mailman/listinfo/haskell-cafe
>