Posted By

rtperson on 06/01/11


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fizzbuzz


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FizzBuzz in Haskell


 / Published in: Haskell
 

The dread FizzBuzz question -- really, a test if your average programmer knows his or her FOR loops. The spec is that you're counting from 1 to 100. Your program should print out "fizz" if the index is divisible by three, and "buzz" if it's divisible by five, and "fizzbuzz" if it's divisible by both.

I have a couple different versions in this code: first, a few functions just fizzing or buzzing. Second, a generalization which allows any standard message against any divisor. Third, a purely functional version that zips two lists together (giving us free concatenation for "fizzbuzz"). Fourth, a list comprehension and lastly, a monadic version that calls a pure function that uses guards.

  1. -- file: Fizz.hs
  2. -- a Haskell implementation of the fizzbuzz problem
  3.  
  4. ns = [0..100] :: [Int]
  5.  
  6. fizz :: Int -> String
  7. fizz n = if (n `mod` 3) == 0 then "fizz" else ""
  8.  
  9. buzz :: Int -> String
  10. buzz n = if (n `mod` 5) == 0 then "buzz" else ""
  11.  
  12. -- a generalized version. It takes the index and the value to divide against, and
  13. -- returns the message if n is evenly divisible by x
  14. fluff :: Int -> Int -> String -> String
  15. fluff n x message = if (n `mod` x) == 0 then message else ""
  16.  
  17. -- a purely functional implementation.
  18. fizzBuzz :: [String]
  19. fizzBuzz = zipWith (++) (map fizz ns) (map buzz ns)
  20.  
  21. -- another purely functional version. Very easy to remember.
  22.  
  23. threes :: [String]
  24. threes = cycle ["", "", "Fizz"]
  25.  
  26. fives :: [String]
  27. fives = cycle ["", "", "", "", "Buzz"]
  28.  
  29. fizzBuzzCycle :: [String]
  30. fizzBuzzCycle = zipWith (++) threes fives
  31.  
  32. -- List comprehensions, anyone?
  33. boomBang :: [String]
  34. boomBang =
  35. [ if x `mod` 15 == 0
  36. then "boombang"
  37. else if x `mod` 3 == 0
  38. then "boom"
  39. else if x `mod` 5 == 0
  40. then "bang"
  41. else show x
  42. | x <- ns]
  43.  
  44. -- the answer your recruiter is probably looking for
  45. -- (if your recruiter has enough doubts about your
  46. -- programming ability that he/she busts out
  47. -- fizzbuzz on your butt)
  48. main :: IO ()
  49. main = printAll $ map fizz' [1..100]
  50. where
  51. printAll [] = return ()
  52. printAll (x:xs) = putStrLn x >> printAll xs
  53.  
  54. fizz' :: Int -> String
  55. fizz' n
  56. | n `mod` 15 == 0 = "fizzbuzz"
  57. | n `mod` 3 == 0 = "fizz"
  58. | n `mod` 5 == 0 = "buzz"
  59.  
  60. -- or, to get rid of the explicit recursion in the main routine
  61. main2 :: IO ()
  62. main2 = printAll $ map fizz' ns
  63. where
  64. printAll xs = foldr ((>>) . putStrLn) (return ()) xs

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