1. Chapter 7 Logic Instructions and Programs

2. Sections 7.1 Logic and compare instructions
7.2 Rotate and swap instructions
7.3 BCD and ASCII application programs

3. Objective
介紹關於邏輯運算的指令。如ANL、ORL、XRL、CPL等指令。另外有執行 byte 旋轉的指令。如RR、RL、SWAP，簡單可是很有用。
通常這些指令是用於 bit manipulation，我們關心的只是 byte 中的某幾個 bits 而已。
最後有一個範例是利用這些指令做 BCD 與 ASCII 之間的轉換。
我們將只是很簡單的介紹這些指令，如果你想得到更多關於這些指令的用法與範例，如更多的 addressing mode 的用法，請看 Appendix A.1。

4. Section 7.1 Logic and Compare Instructions

5. ANL ANL destination-byte,source-byte MOV A,#35H ;0010 0101
ANL A,#0FH ; => A=
No effect on any of the flags.
ANL is often used to mask (set to 0) certain bits of an operands. X Y
X AND Y 1
AND 2 bits X and Y

6. Example 7-1 Show the results of the following. MOV A,#35H ;A = 35H
ANL A,#0FH ;A = A AND 0FH (now A = 05)
Solution:
35H
0FH
05H H AND 0FH = 05H

7. ORL ORL destination-byte,source-byte MOV A,#35H ;0010 0101
ORL A,#0FH ; => A=
No effect on any of the flags.
ORL is often used to set certain bits of an operands to 1. X Y
X OR Y 1
OR 2 bits X and Y

8. Example 7-2 Show the results of the following. MOV A,#04 ;A = 04
ORL A,#68H ;new A = 6C
Solution:
04H
68H
6CH OR 68 = 6CH

9. XRL ORL destination-byte,source-byte MOV A,#35H ;0010 0101
XRL A,#0FH ; => A=
No effect on any of the flags.
XRL is often used to clear a register, to see if two registers have the same value or to toggle bits of an operands.
unchanged toggled A=
A XRL A => clear A  A=
A XRL B => if A=B then A=0 , the 8051 has no the zero flag.
A XRL with 00H => unchanged  A=
A XRL with FFH => toggle all bits of A  A= X Y
X XOR Y 1
XOR 2 bits X and Y

10. Example 7-3 Show the results of the following. MOV A,#54H XRL A,#78H
Solution:
54H
78H
2CH H XOR 78H = 2CH

11. Example 7-4
The XRL instruction can be used to clear the contents of a register
by XORing it with itself. Show how “XRL A,A” clears A, assuming that A= 45H.
Solution:
45H
XOR a number with itself = 0

12. Example 7-5
Read and test P1 to see whether it has the value 45H. If it does,
send 99H to P2; otherwise, it stays cleared.
Solution:
MOV P2,#00 ;clear P2
MOV P1,#0FFH ;make P1 an input port
MOV R3,#45H ;R3 = 45H
MOV A,P1 ;read p1
XRL A,R3
JNZ EXIT ;jump if A ≠ 0
MOV P2,#99H
EXIT: ...

13. CPL (Complement Accumulator)
CPL A
MOV A,#55H ;
CPL A ;
No effect on any of the flags.
This is also called 1’s complement.

14. Example 7-6 Find the 2’s complement of the value 85H. Solution:
MOV A,#85H ; H =
CPL A ;1’s comp. 1’s =
ADD A,#1 ;2’s comp
= 7BH

15. CJNE (1/2) Compare and Jump if Not Equal
CJNE destination, source, relative address
MOV A,#55H
CJNE A,#99H,NEXT
;do here if A=99H
NEXT: ;jump here if A99H
The compare instructions really a subtraction, except that the operands themselves remain unchanged.
Flags are changed according to the execution of the SUBB instruction.

16. CJNE (2/2) This instruction affects the carry flag only.
CJNE R5,#80,NEXT
;do here if R5=80
NEXT: JNC LAR
;do here if R5>80
LAR: ;do here if R5<80
Compare
Carry Flag
destination > source
CY = 0
CY = 1
Table 7-1

17. Example 7-7
Examine the following code, then answer the following questions.
Will it jump to NEXT?
What is in A after the CJNE instruction is executed?
MOV A,#55H
CJNE A,#99H,NEXT
...
NEXT: ...
Solution:
Yes, it jumps because 55H and 99H are not equal.
A = 55H, its original value before the comparison.

18. Example 7-8
Write code to determine if register A contains the value 99H. If so,
make R1 = FFH; otherwise, make R1 = 0.
Solution:
MOV R1,# ;clear R1
CJNE A,#99H,NEXT ;if A≠99, then jump
MOV R1,#0FFH ;if A＝99, R1=FFH
NEXT: ;if A≠99, R1=0
OVER:...

19. Example 7-9
Assume that P1 is an input port connected to a temperature sensor.
Write a program to read the temperature and test it for the value 75.
According to the test results, place the temperature value into the
registers indicated by the following.
If T = then A = 75
If T < then R1 = T
If T > then R2 = T
Solution:
MOV P1,#0FFH ;make P1 an input port
MOV A,P ;read P1 port
CJNE A,#75,OVER ;jump if A≠75
SJMP EXIT ;A=75
OVER: JNC NEXT ;
MOV R1,A ;A<75, save A R1
SJMP EXIT ;
NEXT: MOV R2,A ;A>75, save A in R2
EXIT: ...

20. Example 7-10
Write a program to monitor P1 continuously for the value 63H. It
should get out of the monitoring only if P1 = 63H.
Solution:
MOV P1,#0FFH ;make P1 an input port
HERE:MOV A,P ;get P1
CJNE A,#63,HERE ;keep monitoring unless
; P1=63H

21. Example 7-11
Assume internal RAM memory locations 40H – 44H contain the
daily temperature for five days, as shown below. Search to see if any
of the values equals 65. If value 65 does exist in the table, give its
location to R4; otherwise, make R4 = 0.
40H=（76）41H=（79）42H=（69）43H=（65）44H=（62）
Solution:
MOV R4,#0 ;R4=0
MOV R0,#40H ;load pointer
MOV R2,#05 ;load counter
MOV A,# ;A=65, value searched for
BACK:CJNE RAM data with 65
MOV R4,R0 ;if 65, save address
SJMP EXIT ;and exit
NEXT:INC R ;increment pointer
DJNZ R2,BACK ;keep check until count=0
EXIT ...

22. Section 7.2 Rotate and Swap Instructions

23. RR (Rotate A Right) RR A MSB LSB MOV A,#36H ;A=0011 0110
RR A ;A=
RR A ;A=
RR A ;A=
RR A ;A=
MSB LSB

24. RL (Rotate A Left) RL A MSB LSB MOV A,#36H ;A=0011 0110
RL A ;A=
RL A ;A=
RL A ;A=
RL A ;A=
MSB LSB

25. RR (Rotate A Right Through Carry)
RRC A
MOV A,#36H ;A= , CY=0
RRC A ;A= , CY=0
RRC A ;A= , CY=1
RRC A ;A= , CY=1
RRC A ;A= , CY=0
MSB LSB CY

26. RLC (Rotate A Left Through Carry)
RLC A
MOV A,#36H ;A= , CY=1
RLC A ;A= , CY=0
RLC A ;A= , CY=0
RLC A ;A= , CY=1
RLC A ;A= , CY=1 CY
MSB LSB

27. SWAP A SWAP A before: before: 0111 0010 after: after: 0010 0111
MOV A,#72H ;A=72H
SWAP A ;A=27H
before:
D7 – D4
D3 – D0
before:
0111
0010
after:
D3 – D0
D7 – D4
after:
0010
0111

28. Example 7-12
(a) Find the contents of register A in the following code.
(b) In the absence of a SWAP instruction, how would you
exchange the nibbles? Write a simple program to show the process.
Solution:
(a)
MOV A,#72H ;A = 72H
SWAP A ;A = 27H
(b)
MOV A,#72H ;A=
RL A ;A=
RL A ;A=
RL A ;A=
RL A ;A=

29. Example 7-13
Write a program that finds the number of 1s in a given byte.
Solution:
MOV R1,#0 ;R1 keeps the number of 1s
MOV R7,#8 ;counter=08 rotate 9 times
MOV A,#97H ;find the # of 1s in 97H
AGAIN:RLC A ;rotate it through the CY
JNC NEXT ;check for CY
INC R ;if CY=1 then add R1
NEXT: DJZN R7,AGAIN;go through this 8 times

30. Example Of Serial Communication
Write a program to transfer data to serial memories such as serial EEPROMs.
Solution:
...
RLC A ;first bit to carry
MOV P1.3,C ;output carry as data bit
RLC A ;second bit to carry
MOV P1.3,C ;third carry as data bit

31. Section 7.3 BCD and ASCII Application Programs

32. Table 7-2. ASCII Code for Digits 0 – 9
Key
ASCII (hex)
Binary
BCD (unpacked) 30 1 31 2 32 3 33 4 34 5 35 6 36 7 37 8 38 9 39

33. Conversion of BCD and ASCII
There is a real time clock (RTC) in many new microcontrollers.
Ex: DS5000T has RTC
RTC keep the time (hour, minute, second) and date (year, month, day) when the power is off.
This data is provided in packed BCD.
For this data to be displayed (ex: on an LCD), it must be in ASCII format.
We show above instructions in the conversion of BCD and ASCII

34. Packed BCD to ASCII Conversion
To convert packed BCD to ASCII
It must be converted to unpacked BCD first.
MOV A,#29H
ANL A,#0FH ;get the lower nibble
The unpacked BCD is tagged with 30H
ORL A,#30H ;make it an ASCII,A=39H ‘9’

35. Example 7-14 (modified)
Assume that register A has packed BCD, write a program to convert
packed BCD to two ASCII numbers and place them in R2 and R6.
Solution:
MOV A,#29H ;packed BCD
ANL A,#0FH ;Lower nibble: A=09H
ORL A,#30H ;make it an ASCII, A=39H (‘9’)
MOV R6,A ;R6=39H ASCII char
MOV A,#29H ;
ANL A,#0F0H ;upper nibble: A=20H
SWAP A ;A=02H, equals to ”RR A” 4 times
ORL A,#30H ;A=32H,ASCII char.’2’
MOV R2,A ;R2=32H ASCII char

36. ASCII to packed BCD Conversion
To convert ASCII to packed BCD
It must be converted to unpacked BCD first.
MOV A,#’2’ ;A=32H
ANL A,#0FH ;get the lower nibble
MOV R1,#’9’ ;R1=39H
ANL R1,#0FH ;get the lower nibble
Combined them to the packed BCD.
SWAP A ;become upper nibble A=20H
ORL A,R1 ;packed BCD,A=29H

37. You are able to
Define the truth tables for logic functions AND, OR, XOR
Code 8051 Assembly language logic function instructions
Use 8051 logic instructions for bit manipulation
Use compare and jump instructions for program control
Code 8051 rotate and swap instructions
Code 8051 programs for ASCII and BCD data conversion

38. Homework Chapter 7 Problems：4,5,10,11,14,15 Note:
Please write and compile the program of Problems 10,11,14,15