175
In the case below, A6E2 + 80C5 = 127A7. The result is a 5-digit number, so
CY (SR 25504) = 1, and the content of R + 1 becomes #0001.
R+1: HR 11
R: HR 10
0
0
0
1
2
7
A
7
Au: IR 010
A
6
E
2
Ad: DM 0100
8
0
C
5
+
The following example performs eight-digit addition by using ADB(50) twice.
ADB(50) is also used to place the carry into DM 0302 (one word greater than
the rest of the answer). The complete answer thus ends up in DM 0300
through DM 0302.
CLC(41)
00000
@ADB(50)
LR 20
DM 0200
DM 0300
@ADB(50)
LR 21
DM 0201
DM 0301
@ADB(50)
#0000
#0000
DM 0302
Address
Instruction
Operands
00000
LD
00000
00001
CLC(41)
00002
@ADB(50)
LR
20
DM
0200
DM
0300
00003
@ADB(50)
LR
21
DM
0201
DM
0301
00004
@ADB(50)
#
0000
#
0000
DM
0302
In the case below, 4F EC3B80C5 = 13B8E27A7. The sum of the
addition of the lower 4 digits is a 5-digit number, so CY (SR 25504) = 1, and
the sum of the higher 4-digit addition is incremented by 1.
R: DM 0300
2
7
A
7
Au: LR 20
A
6
E
2
Ad: DM 0200
8
0
C
5
+
R: DM 0301
3
B
8
E
Au: LR 21
4
F
5
2
Ad: DM 0201
E
C
3
B
+
CY = 1
Lower 4 digits.
Higher 4 digits.
CY = 1
R+2: DM 0302
R+1: DM 0301
R: DM 0300
0
0
0
1
3
B
8
E
2
7
A
7
Binary Calculations
Section 5-18
