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at August 4, 2012 01:08 by rtperson
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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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at June 1, 2011 00:14 by rtperson
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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. Look, no monads! fizzBuzz :: [String] fizzBuzz = zipWith (++) (map fizz ns) (map buzz ns) -- 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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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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FizzBuzz in Haskell
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Haskell