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

[Donn Cave]

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.

	Donn