## A CCR*

## num_1: dc.b $40

## num_2: dc.b $50

## start: ldaa num_1 40 ....

## adda num_2 90 N.V.

## * bits 0-3 only, āNā means N bit (bit 3) is 1

## $40 = % 0100 0000 (carry from bit 6 to bit 7 so the sum,

## + $

__50__= %__0101 0000__viewed in 2's comp. notation,## $90 = % 1001 0000 of these two pos. numbers is neg.)

## Of course, what has happened is that the sum $90 = 144 > 127.