Learn You a Haskell shows the lines function:

It takes a string and returns every line of that string in a separate list.



Here's my implementation:

lines' :: String -> [String]
lines' [] = []
lines' xs = lines'' xs []
             where lines'' [] ys = ys : []
                   lines'' (x:xs) ys 
                     |  x == '\n' = ys : lines'' xs []
                     | otherwise  = lines'' xs (ys ++ [x])

Please critique it. Also, when using an accumulator value (such as ys in lines''), I don't know how to use the : function instead of ++.


1 Answer 1

  1. What happens when a string ends in a newline? lines "\n"[""], but lines' "\n"["", ""].

  2. ys could have a more informative name. line?

  3. When you find yourself wanting to add new items to the end of a list, consider keeping the list in reverse order, adding items to the front and reversing it when it's done. This avoids needing to use ++. ("Done" is a fuzzy concept when working lazily with infinite lists.)

  4. ...but when reinventing the wheel, you needn't reinvent the axle and bearings as well. There is a Prelude function that will parse one line from a string: break. If you use that, you need only handle recursing on the remainder. This is how lines is really implemented.

    lines s         =  let (l, s') = break (== '\n') s
                       in  l : case s' of
                                    []      -> []
                                    (_:s'') -> lines s''
  • \$\begingroup\$ thanks, Anonymous. Excellent suggestions for improvement. Would you implement #1 by simply adding that special case lines' "\n" = [""]? \$\endgroup\$ May 13, 2014 at 23:47
  • \$\begingroup\$ No, because it wouldn't cover other strings ending in newlines (e.g. lines "a\n"["a"]), and because it can be done without a special case by recursing appropriately. \$\endgroup\$
    – Anonymous
    May 14, 2014 at 0:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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