FizzBuzz up to 99 in Haskell

Question

This code solves the problem of FizzBuzz. Is it possible in any way to improve it?

main = main' 1 where
main' n = do
 (putStrLn . choose) (show n, "Fizz", "Buzz", "FizzBuzz", n)
 if n < 100 then main' (succ n) else putStrLn "End!" 
 where
 choose (n0, n3, n5, n15, n) 
 | mod n 3 == 0 && mod n 5 == 0 = n15
 | mod n 5 == 0 = n5
 | mod n 3 == 0 = n3
 | True = n0

• jnewman: Which definition of FizzBuzz are you using? Based on the first Google result the #s are fixed at [1..100], so there shouldn't be any inputs.

Answer 1

You could separate your I/O from the pure code:

fizzBuzz :: Int -> String
fizzBuzz n | mod n 3 == 0 && mod n 5 == 0 = "FizzBuzz"
 | mod n 5 == 0 = "Buzz"
 | mod n 3 == 0 = "Fizz"
 | otherwise = show n
main = mapM print (map fizzBuzz [0..100])

• Komi Golov: I think mapM (print . fizzBuzz) [0..100] would be better.
• hammar: mapM_ would be even better.

Answer 2

jaket is definitely right, the pure/impure distinction is important. My addition would be that you should avoid recomputing the modulo:

fizzBuzz :: Int -> String
fizzBuzz n | fizz && buzz = "FizzBuzz"
 | buzz = "Buzz"
 | fizz = "Fizz"
 | otherwise = show n
 where fizz = mod n 3 == 0
 buzz = mod n 5 == 0
sfb = map fizzBuzz [1..15]

Answer 3

I had written something similar to Will, but the spec I read here says, that FizzBuzz should always cover [1..100], so my implementation was a bit different:

show' :: Int -> String
show' n
 | fizz && buzz = "FizzBuzz" 
 | buzz = "Buzz"
 | fizz = "Fizz"
 | otherwise = show n
 where fizz = mod n 3 == 0
 buzz = mod n 5 == 0
fizzBuzz = [show' x | x <- [1..100]]

Will's idea about using where to cache the mod result was a nice idea IMHO.