Operator MOD

[SMcNeill]

As I posted for Pete -- getting the pattern one wants is completely trivial.  Why is this such an issue for folks??

Code: (Select All)

For i = 10 To -10 Step -1
    Print i Mod 5, ModX(i, 5)
Next

Function ModX (num1, num2)
    ModX = ((num1 Mod num2) + num2) Mod num2
End Function

^That's it.  That's all one has to do to get the results they want.   If you have some God-fearing aversion for Functions, then just use  ((num1 Mod num2) + num2) Mod num2 directly.   It's not rocket science -- it's just remainders after division!