[Haskell-cafe] Separate a string into a list of strings

[Clifford Beshers]

Donn Cave wrote:
> Quoth Clifford Beshers <clifford.beshers at linspire.com>:
> | Well, I couldn't resist the puzzle.  Here are solutions using foldr and 
> | unfoldr.  Don't know if they are cunning or not, but they were kind of fun.
>
> ...
> | splitByElem1 e xs =
> |     foldr f [[]] xs
> |     where f a b = if a == e then [] : b else (a : head b) : (tail b)
>
> This does the right thing with trailing separators, which is not to be
> taken for granted among Haskell split implementations.  The splits I have
> been seeing in this thread are conservative, so if the separator is ':',
> then "::a" splits to ["", "", "a"].  Frequently however the implementation
> fails to deal with the trailing separator, so "a:" is ["a"], where it
> should be ["a", ""].  It's not something you run into right away.
>
> In a liberal split, "a " should indeed be ["a"], but that's a different
> matter.  Neither of the two I've looked at seems to be shooting for a
> liberal "words" white space split.
Good point.  My solutions are inconsistent on white space, which I don't 
like.

Your criterion eliminates this solution as well:

filter (/= ",") $ groupBy (\x y -> x /= ',' && y /= ',') "Haskell, 
Haskell, and Haskell,"
["Haskell"," Haskell"," and Haskell"]