3. Instruction Set
3-233
4. When n = 4, the bit structure will be as:
0100 1 0 1 110000000
0 00011 01 01011000
10011 11011110111
CDE FD10 D20D21
44H D
46H F
43H C
45H E
5. When n = 1 ~ 16:
D
n D13 D12 D11 D10
1 ***C H
2 **CD H
3 *CDE H
4
CDEF H
5 ***C H DEF8 H
6 **CD H EF89 H
7 *CDE H F89A H
8
CDEF H 89AB H
9 ***C H DEF8 H 9AB4 H
10 **CD H EF89 H AB45 H
11 *CDE H F89A H B456 H
12
The
undesignated
parts in the
registers in use
are all 0.
CDEF H 89AB H 4567 H
13 ***C H DEF8 H 9AB4 H 5670 H
14 **CD H EF89 H AB45 H 6701 H
15 *CDE H F89A H B456 H 7012 H
16 CDEF H 89AB H 4567 H 0123 H
Program Example 2:
1. M1161 = ON: 8-bit conversion.
X0 HEX D20 D10 K4
M1000 M1161
2. Assume:
S ASCII code HEX
conversion S ASCII code HEX
conversion
D20 H 43 “C” D25 H 39 “9”
D21 H 44 “D” D26 H 41 “A”
D22 H 45 “E” D27 H 42 “B”
D23 H 46 “F” D28 H 34 “4”