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Fizz Buzz in Haskell
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gistfile1.hs
-- Feels like there's a shorthand for the duplicated `mod` expressions.
fizzBuzz x
| (x `mod` 3) == 0 && (x `mod` 5) == 0 = "FizzBuzz"
| (x `mod` 3) == 0 = "Fizz"
| (x `mod` 5) == 0 = "Buzz"
| otherwise = show x
-- Extracting the mod expressions in any way I know how (so far) leads to more complication.
@srbaker
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srbaker commented May 7, 2014

I also hate that it's damned order dependent.

@ukd1
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ukd1 commented May 7, 2014

Nice, it looks clear! Dumb q, but can't you do x mod 15 for FizzBuzz?

@srbaker
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srbaker commented May 7, 2014

Ah yes, I had 15 originally for brevity, but wasn't sure if it applied completely. It does.

@srbaker
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srbaker commented May 7, 2014

I'm not there yet, but I'm pretty sure I can just do a lookup table mapping 3:fizz,5:buzz,15:fizzbuzz

@jbrains
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jbrains commented May 7, 2014

Guard Clauses depend on sequence. Always has been; always will be.

@jbrains
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jbrains commented May 7, 2014

I think this is about as good as mine's going to get: https://gist.github.com/jbrains/2a29465457bddcca0c45

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