3. Instruction Set
3-281
4. When n is 2, the bit structure will be as:
0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 1
0 1
2 3
D10=0123 H
0 0 0 0 0 0 0 1 1 0 0 0
0 0 0 0 0 0 1 1 0 0 1
3
3 3
2
10 0 0
10 0 0 0
ASCII code of "2" in D20 is 32H
ASCII code of "3" in D21 is 33H
5. When n is 4, the bit structure will be as:
0 0 0 0 0 1 0 1 11 0 0 0000
0 0 0 0 0 0 0 0 0 0 0 00 0 1 1
0 0
0 1 2 3
D10 = H 0123
b15
b15
Converted to
b15
0 0 0 0 0 0 0 0 0 0 1 10 0 1 1
b15
3 H 33
2 H 32
D22
b15
b0
b0
b0
b0
b0
D20
D21
1 H 31
D23
0 H 30
0 0 0 0 0 0 0 0 0 0 10 1 1
0 0
0 0 0 0 0 0 0 0 0 1 00 1 1
6. When n = 1 ~ 16:
K1 K2 K3 K4 K5 K6 K7 K8
D23
“3” “2” “1” “0” “7”
No
change
D30
D33