14 Sep
2016
14 Sep
'16
7:42 a.m.
On 14/09/2016 13:17, Vincent Slyngstad wrote:
From: Kyle Owen: Tuesday, September 13, 2016 7:12 PM
Yes, the usual way is just what you describe: A + B - (2 * A AND B),
which works by calculating the bitwise carries and subtracting them from
the addition, to make it carryless. The AND tells you where they result
TAD OVFA /A XOR B
AND OVFB
CMA IAC
TAD OVFA
TAD OVFB
Sigh.
There seems to be an issue with my implementation of XOR.
Before the CMA IAC there needs to be CLL RAL, to reposition the carries
before subtracting them. Both XORs are affected. from but you have to shift them to the correct
positions before taking
the difference. You were missing the multiply x 2 (which
needs the CLL
first). Easily missed.
But unless I've missed something, you don't need a complete XOR, you
just need to check the sign bits.
--
Pete
Pete Turnbull