5 ms·

How can FizzBuzz be complicated for anybody? The solution is so obvious. I thought it was some sort of trick question but apparently not. http://wiki.c2.com/?F

by rum3 7y ago

How can FizzBuzz be complicated for anybody? The solution is so obvious. I thought it was some sort of trick question but apparently not. http://wiki.c2.com/?FizzBuzzTest http://wiki.c2.com/?FizzBuzzTest

k_sze 7y ago
You'd be amazed at the percentage of people who fail this test. I can't say I have interviewed many people, but even seeing one or two people fail this test for a technical role is mind-boggling.
- james_s_tayler 7y ago
  I could imagine someone not from a CS background to fail it by perhaps not recognizing they need to the modulus operator or even being aware of its existence? Outside that, yeah. It indeed boggles the mind that the vanilla version of it could be failed by anyone who has taken elementary programming course.
  - lincpa 7y ago
    In order to test your real problem-solving skills, you should allow the use of Internet search, because not all computer professionals can remember a particular algorithm.
james_s_tayler 7y ago
Well... vanilla FizzBuzz is easy peasy. Ask them to draw an architecture diagram for FizzBuzz Enterprise Edition on a whiteboard. Or ask them to implement it without using the modulus operator... muahaha. I'm sure that'd stump almost everyone. You can make FizzBuzz incredibly complicated if you deviate from the vanilla FizzBuzz.
- billpg 7y ago
  My favorite is the variation where the base numbers and replacement strings are supplied as run-time parameters. RunFizzBuzz(""); RunFizzBuzz("3=Fizz,5=Buzz"); RunFizzBuzz("7=Rutabaga");
- rum3 7y ago
  Asking for it without using the modulus operator is a bit evil haha. I like this exercise and I sent it to a friend who has a few years of programming experience but is still learning heavily. The first solution lacked the number being printed but everything else was right. Second solution had the number, but he missed the FizzBall when the number was divisible by both three and five. The third solution lacked the Fizz when it was divisible by three. Finally his fourth solution was correct.

by rum3 7y ago

k_sze 7y ago

You'd be amazed at the percentage of people who fail this test. I can't say I have interviewed many people, but even seeing one or two people fail this test for a technical role is mind-boggling.

james_s_tayler 7y ago
I could imagine someone not from a CS background to fail it by perhaps not recognizing they need to the modulus operator or even being aware of its existence? Outside that, yeah. It indeed boggles the mind that the vanilla version of it could be failed by anyone who has taken elementary programming course.
- lincpa 7y ago
  In order to test your real problem-solving skills, you should allow the use of Internet search, because not all computer professionals can remember a particular algorithm.

james_s_tayler 7y ago

Well... vanilla FizzBuzz is easy peasy. Ask them to draw an architecture diagram for FizzBuzz Enterprise Edition on a whiteboard. Or ask them to implement it without using the modulus operator... muahaha. I'm sure that'd stump almost everyone. You can make FizzBuzz incredibly complicated if you deviate from the vanilla FizzBuzz.

billpg 7y ago
My favorite is the variation where the base numbers and replacement strings are supplied as run-time parameters. RunFizzBuzz(""); RunFizzBuzz("3=Fizz,5=Buzz"); RunFizzBuzz("7=Rutabaga");
rum3 7y ago
Asking for it without using the modulus operator is a bit evil haha. I like this exercise and I sent it to a friend who has a few years of programming experience but is still learning heavily. The first solution lacked the number being printed but everything else was right. Second solution had the number, but he missed the FizzBall when the number was divisible by both three and five. The third solution lacked the Fizz when it was divisible by three. Finally his fourth solution was correct.