Take the 2-minute tour ×
Code Review Stack Exchange is a question and answer site for peer programmer code reviews. It's 100% free, no registration required.

I implemented the split function in Haskell:

split :: (Eq a) => a -> [a] -> [[a]]
split _ [] = []
split x ys = f : split x rest 
  where (f, rest) = break (== x) (dropWhile (== x) ys) 

Note that I'm calling dropWhile (== x) since break's second-tuple value will include the "broken on" value.

example:

*Main> break (== 'a') "dogacactus"
("dog","acactus")

Testing

*Main> split2 '3' "123aaaBBB3"
["12","aaaBBB",""]
share|improve this question

1 Answer 1

up vote 6 down vote accepted

Generally it looks good, at least to an Haskell beginner like me. I find the use of dropWhile to be a bit confusing. I admit I had to read your comment to understand why you used it.

I think it could also have introduced a bug. What is the expected behaviour for split ',' ",,"? Your code returns [""] but I would expect ["","",""]. Was your behaviour intended?

This solution fixed the issue.

split :: (Eq a) => a -> [a] -> [[a]]
split _ [] = []
split separator ys = f : (split separator (dropSeparator separator rest))
  where (f, rest) = break (== separator) ys

dropSeparator :: Eq a => a ->  [a] -> [a]
dropSeparator _ [] = []
dropSeparator separator (x:xs) = if x == separator then xs else x:xs

I prefer using meaningful variable names, such as separator, even if I'm not sure if it is idiomatic Haskell. Ditto for the parentheses and the introduction of the helper function.

share|improve this answer
    
thank you for letting me know that my usage of dropWhile was both confusing and incorrect! Initially I wanted to do something like where (f, _:rest) = ... , but that would fail to match for []. –  Kevin Meredith Sep 16 '14 at 20:35

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.