[XorNot]

Do you tell them what the "mod" operator is before giving it?

The failure rate of FizzBuzz has always struck me as depending on the idea that you can do a lot of programming and just never need that operator.

[arethuza]

I once worked with a guy who was an incredibly good developer and I was surprised when he didn't see anything special about the number 64 (i.e. a power of two) - turns out that he'd never done any bit fiddling type work so he hadn't had to think in those terms. It wouldn't surprise me if a lot of people hadn't heard of "mod" either....

[dkdbejwi383]

A huge majority of programming work is basically just CRUD stuff and other data shuffling. It’s not surprising that someone wouldn’t have needed to work with big shifting (or modulus) in that case.

[arethuza]

He was an expert in complex multi-organisation enterprise integration and was the go to guy to work out why horrific distributed transactions were failing... He also did a lot of cool stuff as side projects in his own time - just none of them happened to involve worrying about powers of 2.

[azornathogron]

You don't actually need mod to do fizzbuzz, even if that's the most obvious way for people who know what mod is.

But without any "real" math at all you can do it with, eg, two counters and some if statements. Or if you recognise that there's a repeating pattern you can work out that pattern manually and just write code to emit it over and over.

[eschneider]

Even if that's so, modulus (or at least the concept of remainders) are elementary school math and any competent programmer could bang together an (inefficient) modulus operator in a few minutes.

So even in a language w/o a mod operator, it's not a hard problem if you understand how to solve problems with code.

[daemin]

Unless you specifically want a compiling and running version of FizzBuzz you don't actually need to use or know about the mod operator.

At least for me it would be sufficient if the person used a function like IsMultipleOf(x, m), or Remainder(x, n). This would at least make it clear what the function did even if they didn't get the exact operator.

The other thing to note is that the mod operator works differently on different languages and platforms.

[DonHopkins]

> the mod operator works differently on different languages and platforms

Not with positive inputs, which is the domain of FizzBuzz.

Even if you don't know what "mod" means, if you have no idea know what a remainder is, and that the problem calls for it, and you can't derive the mod operator using integer addition, subtraction, multiplication, and division, then your math and problem solving skills are pretty weak, which FizzBuzz tests.

[sjaak]

No but I permit internet access as long as they don't search for the solution (I trust them not to do that, and don't monitor what they do)

[DonHopkins]

You're too trusting.

[ownagefool]

Whilst I agree, his point still stands that either they do it fast or never.

[ownagefool]

I stopped using fizz buzz a long time ago. 90% of candidates can't define a 2d array in their chosen language without first filtering the candidates, where you get to about 50%.

[sensanaty]

Yeah I know how to implement FizzBuzz since it's such a meme, but I've basically never used the mod operator in real code. Maybe it comes up in more math-y code I suppose, but for most backend/frontend/SQL code I've never reached for it.

[wruza]

  for (…) {
    heavy_op();
    if (i % 100 == 0) {
      printf("not dead");
    }

Classic

[stefan_]

More mathy code like checking if a number is even or splitting a total number of seconds into minutes and seconds?

[vundercind]

I’ve used it for coloring alternating lines differently in UI code, and as a lazy way to log only every so many times some loop runs.

I only know it well because it was covered near the beginning of one of the first programming books I picked up (on Perl 5) and it stuck with me because it seemed wild to me that they had an operator for that.

[kmoser]

Depends on the application. I've written accounting software that makes use of it, along with heavy use of floor() and ceil(), including in SQL.