The problem is here, the solutions are here. I don't see a solution like mine (maybe the 4th), it seems to work though.

-- ---------------------------
-- 18 (**) Extract a slice from a list (idx start from 1)

slice _      _ 0 = []
slice (x:xs) 1 k = x : slice xs 1 (k-1)
slice (x:xs) i k = slice xs (i-1) (k-1)

I tried testing it with QuickCheck against the drop/take solution, but it gives up because of the very restrictive conditions (1 <= i <= k <= len xs) and I don't want to get deeper in testing right now.

Is this good Haskell code? Am I forgetting any edge case? (...should I suggest it on the solutions wiki?)


Yes you forgot to cover some cases.

Your compiler should be able to detect one of them. Here is what ghc -Wall says:

slice.hs:2:1: Warning:
    Pattern match(es) are non-exhaustive
    In an equation for `slice':
        Patterns not matched: [] _ (GHC.Types.I# #x) with #x `notElem` [0#]

slice.hs:4:8: Warning: Defined but not used: `x'

This version fixes those two warnings:

slice []     _ _ = []
slice _      _ 0 = []
slice (x:xs) 1 k = x : slice xs 1 (k-1)
slice (_:xs) i k = slice xs (i-1) (k-1)

But it is still incorrect, because there is another case that the compiler doesn't warn you about: a negative k value. Your code will return the entire list instead of an empty list.

Consequently, if you use a negative k on an infinite list you get an infinite loop:

> print $ slice [1..] 1 (-1)
  • \$\begingroup\$ Very exhaustive (negative i too is a problem). Indeed, I supposed strictly valid input. +1 also for the -Wall suggestion (which told me about other bad habits I was developing). \$\endgroup\$
    – bigstones
    Aug 29 '13 at 15:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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