Edit: the second TL;DR point should instead read “use this formula to correct for negative divident”. Remember, this fails with something as basic as (3 % 10), but can be used to correct for negative dividents in your error handling.
Nice, I was able to make an “Absolute Modulo” function in my function library to account for this. Thanks for the answer!
b-Abs(a%b)
Isn’t it easier to say a+b in case of a negative number?
In the example:
- b-Abs(a%b) = 4 - (1%4) = 3
- a+b = -1 + 4 = 3
Nice, thanks for coming up with the calculation.
Hi all,
Cool post @Dextrices
I have a ‘snippet’ for this in my library. I feel like it is relevant. 
Cheers
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You will need to create function which in case of negative…
@Shadowriver, what would be a good name for such a function?