Revision: 47082
Updated Code
at August 4, 2012 01:08 by rtperson
Updated Code
-- 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
Revision: 47081
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at June 1, 2011 00:14 by rtperson
Initial Code
-- 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
Initial URL
Initial Description
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.
Initial Title
FizzBuzz in Haskell
Initial Tags
Initial Language
Haskell