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rtperson on 06/01/11

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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.

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-- file: Fizz.hs
-- a Haskell implementation of the fizzbuzz problem

ns = [0..100] :: [Int]

fizz :: Int -> String
fizz n = if (n `mod` 3) == 0 then "fizz" else ""

buzz :: Int -> String
buzz n = if (n `mod` 5) == 0 then "buzz" else ""

-- a generalized version. It takes the index and the value to divide against, and 
-- returns the message if n is evenly divisible by x
fluff :: Int -> Int -> String -> String
fluff n x message = if (n `mod` x) == 0 then message else ""

-- a purely functional implementation.
fizzBuzz :: [String]
fizzBuzz = zipWith (++) (map fizz ns) (map buzz ns)

-- another purely functional version. Very easy to remember.

threes :: [String]
threes = cycle ["", "", "Fizz"]

fives :: [String]
fives = cycle ["", "", "", "", "Buzz"]

fizzBuzzCycle :: [String]
fizzBuzzCycle = zipWith (++) threes fives

-- List comprehensions, anyone?
boomBang :: [String]
boomBang = 
    [ if x `mod` 15 == 0
        then "boombang" 
        else if x `mod` 3 == 0
                then "boom" 
                else if x `mod` 5 == 0
                    then "bang" 
                    else show x 
      | x <- ns]

-- the answer your recruiter is probably looking for
-- (if your recruiter has enough doubts about your  
-- programming ability that he/she busts out 
-- fizzbuzz on your butt)
main :: IO () 
main = printAll $ map fizz' [1..100]
         where 
            printAll [] = return ()
            printAll (x:xs) = putStrLn x >> printAll xs

fizz' :: Int -> String
fizz' n
    | n `mod` 15 == 0  = "fizzbuzz"
    | n `mod` 3 == 0   = "fizz"
    | n `mod` 5 == 0   = "buzz"
    | otherwise          = show n

-- or, to get rid of the explicit recursion in the main routine 
main2 :: IO ()
main2 = printAll $ map fizz' ns
        where 
            printAll xs = foldr ((>>) . putStrLn) (return ()) xs

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