Oh, and by the way, this is not at all a complaint, I applaude this discrepancy to the manual. The fraction part of a division can easily be computed by
fpart(a/b)
while mirroring the "normal" mod() behavior with a mod function as described in the manual is a tad more irky.
Bye, Malte.
... as any well-behaved mod function should do. But indeed, you are correct in that the documentation is wrong.
~Dave
Just a quick note: other than stated in the help, Inquisit's mod() function computes the integer remainder of a division.Examplesmod(1,2) = 1mod(2,9) = 2mod(11,4) = 3I haven't tested what mod(3.14,2) would yield but I assume it to be 1.14.Bye, Malte