2 ms·
So, instead of checking twice if the number is divisible by five, you check once but then you check the type of every instruction again. How is that a win?
by ohbtvz 4y ago
So, instead of checking twice if the number is divisible by five, you check once but then you check the type of every instruction again. How is that a win?
- quickthrower2 4y agoYeah it is a bit of a strawman, especially as Exhibit B solves this problem (and an inline function would remove the duplication). However as the mention at the end this isn't a serious performance tuning session, or even a serious answer to FizzBuzz, but a way to introduce some Haskell concepts.