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Comment by vessenes | Hacker News Reader

I suppose it's a style issue as to whether an extra mod is better than checking if one of the triggers has passed. I would propose it's more DRY to do it the short way though.

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It's not any shorter. The extra check as to whether or not to print the line break cancels out the mod 15 check. In my opinion, it's cleaner to have three conditionals of the same type than two checking mods and a third checking the OR of the first two.

Of course, it can be actually shorter with a goto.

Whether you print the number, Fizz, Buzz or FizzBuzz you are going output a line break, so I'm not sure what you would be checking for. Output \n unconditionally.

then you'd get \n \n fizz\n \n buzz\n rather than fizz\n buzz\n

The canonical FizzBuzz wants

1\n2\nFizz\n4\nBuzz\n...\nFizzBuzz\n

If you did not print anything for some cases, you're right that it could not be unconditional (though it could still be done at the end with a flag).

oops, yes, my mistake.

Then you call `printf` twice every loop instead of once. `printf` is buffered so you aren't making two system calls, but you are still making two function calls.

Maybe it's a performance/brevity compromise, but the latter is the issue the I addressed. The least-calls solution would probably be to print the whole output as a single string constant.

    string output = numb.toString();
    if(numb % 3  == 0) ouput = "Fizz";
    if(numb % 5  == 0) output.Append("Buzz");
    write output;

It is possible to use two checks to cover all three fizzbuzz components.

5Buzz

In some languages you can hide the extra check:

for i in range(1,100):print("Fizz"(i%3==0) + "Buzz"(i%5==0) or i)

For others who are confused by this syntax, it appears that there are *'s that got eaten and turned into italics.

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