> (x rem n + n) rem n Correct me if I'm wrong, but this is also attractive because it can be applied even to an "actual modulo" operator. That is, because "(x

> (x rem n + n) rem n Correct me if I'm wrong, but this is also attractive because it can be applied even to an "actual modulo" operator. That is, because "(x mod n + n) mod n" has the same behavior as the above, one need not even know which way a particular language's "%" operator behaves if one just always uses this filter.