On Friday, 13 November 2015 at 09:33:51 UTC, John Colvin wrote:
unsigned: f(v) = v mod 2^n - 1
signed: f(v) = ((v + 2^(n-1)) mod (2^n - 1)) - 2^(n-1)
I guess you meant mod 2^n in both cases...
If you look at how Mathematics deals with this issue, there is
simply no signed or unsigned arithmetic modulo n, because they
are exactly the same. There are only separate types in
programming languages because the comparison operators are
defined differently on them.
Mathematicians don't define comparison on modular rings, because
it is not possible to do so in a way that is consistent with the
usual rules anyway (e.g. x+1 > x is always false for some x).
