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Chapter 7 Logic Instructions and Programs

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Sections

7.1 Logic and compare instructions

7.2 Rotate and swap instructions

7.3 BCD and ASCII application programs

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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 。

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Section 7.1Logic and Compare Instructions

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ANL

ANL destination-byte,source-byte MOV A,#35H ;0010 0101

ANL A,#0FH ;0000 1111 => A=0000 0101– 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

0 0 0

0 1 0

1 0 0

1 1 1

AND 2 bits X and Y

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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 0 0 1 1 0 1 0 1

0FH 0 0 0 0 1 1 1 1

05H 0 0 0 0 0 1 0 1 35H AND 0FH = 05H

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ORL

ORL destination-byte,source-byte MOV A,#35H ;0010 0101

ORL A,#0FH ;0000 1111 => A=0010 1111– 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

0 0 0

0 1 1

1 0 1

1 1 1

OR 2 bits X and Y

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Example 7-2

Show the results of the following.

MOV A,#04 ;A = 04

ORL A,#68H ;new A = 6C

Solution:

04H 0000 0100

68H 0110 1000

6CH 0110 1100 04 OR 68 = 6CH

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XRL

ORL destination-byte,source-byte MOV A,#35H ;0010 0101

XRL A,#0FH ;0000 1111 => A=0010 1010– 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.

X Y X XOR Y

0 0 0

0 1 1

1 0 1

1 1 0

XOR 2 bits X and Y

unchanged toggled

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Example 7-3

Show the results of the following.

MOV A,#54H

XRL A,#78H

Solution:

54H 0 1 0 1 0 1 0 0

78H 0 1 1 1 1 0 0 0

2CH 0 0 1 0 1 1 0 0 54H XOR 78H = 2CH

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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 01000101

45H 01000101

00 00000000 XOR a number with itself = 0

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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: ...

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CPL (Complement Accumulator)

CPL A MOV A,#55H ;0101 0101

CPL A ;1010 1010– No effect on any of the flags.

– This is also called 1’s complement.

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Example 7-6

Find the 2’s complement of the value 85H.

Solution:

MOV A,#85H ; 85H = 1000 0101

CPL A ;1’s comp. 1’s = 0111 1010

ADD A,#1 ;2’s comp. + 1

0111 1011 = 7BH

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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 A99H– The compare instructions really a subtraction, except that the operands themselves remain unchanged.

– Flags are changed according to the execution of the SUBB instruction.

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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

destination > source CY = 1Table 7-1

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Example 7-7

Examine the following code, then answer the following questions.

(a) Will it jump to NEXT?

(b) What is in A after the CJNE instruction is executed?

MOV A,#55H

CJNE A,#99H,NEXT

...

NEXT: ...

Solution:

(a) Yes, it jumps because 55H and 99H are not equal.

(b) A = 55H, its original value before the comparison.

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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,#0 ;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:...

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Example 7-9Assume 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 theregisters indicated by the following.

If T = 75 then A = 75

If T < 75 then R1 = T

If T > 75 then R2 = TSolution: MOV P1,#0FFH ;make P1 an input port MOV A,P1 ;read P1 port CJNE A,#75,OVER ;jump if A≠75 SJMP EXIT ;A=75OVER: JNC NEXT ; MOV R1,A ;A<75, save A R1 SJMP EXIT ; NEXT: MOV R2,A ;A>75, save A in R2EXIT: ...

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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,P1 ;get P1

CJNE A,#63,HERE ;keep monitoring unless

; P1=63H

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Example 7-11Assume internal RAM memory locations 40H – 44H contain thedaily temperature for five days, as shown below. Search to see if anyof the values equals 65. If value 65 does exist in the table, give itslocation 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,#65 ;A=65, value searched forBACK:CJNE A,@R0,NEXT;compare RAM data with 65 MOV R4,R0 ;if 65, save address SJMP EXIT ;and exitNEXT:INC R0 ;increment pointer DJNZ R2,BACK ;keep check until count=0EXIT ...

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Section 7.2Rotate and Swap Instructions

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RR (Rotate A Right)

RR A MOV A,#36H ;A=0011 0110

RR A ;A=0001 1011

RR A ;A=1000 1101

RR A ;A=1100 0110RR A ;A=0110 0011

MSB LSB

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RL (Rotate A Left)

RL A MOV A,#36H ;A=0011 0110

RL A ;A=0110 1100

RL A ;A=1101 1000

RL A ;A=1011 0001RL A ;A=0110 0011

MSB LSB

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RR (Rotate A Right Through Carry)

RRC A MOV A,#36H ;A=0011 0110, CY=0

RRC A ;A=0001 1011, CY=0

RRC A ;A=0000 1101, CY=1

RRC A ;A=1000 0110, CY=1RRC A ;A=1100 0011, CY=0

MSB LSB CY

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RLC (Rotate A Left Through Carry)

RLC A MOV A,#36H ;A=0011 0110, CY=1

RLC A ;A=0110 1101, CY=0

RLC A ;A=1101 1010, CY=0

RLC A ;A=1011 0100, CY=1RLC A ;A=1001 1001, CY=1

CY MSB LSB

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SWAP A

SWAP A MOV A,#72H ;A=72HSWAP A ;A=27H

0111 0010

0010 0111

before:

after:

D7 – D4 D3 – D0

D3 – D0 D7 – D4

before:

after:

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Example 7-12

(a) Find the contents of register A in the following code.(b) In the absence of a SWAP instruction, how would youexchange 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=0111 0010 RL A ;A=1110 0100 RL A ;A=1100 1001 RL A ;A=1001 0011 RL A ;A=0010 0111

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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 R1 ;if CY=1 then add R1

NEXT: DJZN R7,AGAIN;go through this 8 times

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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 ;output carry as data bit

RLC A ;first bit to carry

MOV P1.3,C ;third carry as data bit

...

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Section 7.3BCD and ASCII Application Programs

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Table 7-2. ASCII Code for Digits 0 – 9

Key ASCII (hex) Binary BCD (unpacked)0 30 011 0000 0000 00001 31 011 0001 0000 00012 32 011 0010 0000 00103 33 011 0011 0000 00114 34 011 0100 0000 01005 35 011 0101 0000 01016 36 011 0110 0000 01107 37 011 0111 0000 01118 38 011 1000 0000 10009 39 011 1001 0000 1001

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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

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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’

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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

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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

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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

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Homework

• Chapter 7 Problems ： 4,5,10,11,14,15• Note:

– Please write and compile the program of Problems 10,11,14,15