6 arithmetic logic inst and prog

Home Automation, Networking, and Entertainment Lab
Dept. of Computer Science and Information Engineering
National Cheng Kung University, TAIWAN
Chung-Ping Young
楊中平
ARITHMETIC & LOGIC
INSTRUCTIONS AND
PROGRAMS
The 8051 Microcontroller and Embedded
Systems: Using Assembly and C
Mazidi, Mazidi and McKinlay
ARITHMETIC, LOGIC
INSTRUCTIONS AND
PROGRAMS

Department of Computer Science and Information Engineering
National Cheng Kung University, TAIWAN 2
HANEL
ARITHMETIC
INSTRUCTIONS
Addition of
Unsigned
Numbers
ADD A,source ;A = A + source
The instruction ADD is used to add two
operands
Destination operand is always in register A
Source operand can be a register,
immediate data, or in memory
Memory-to-memory arithmetic operations
are never allowed in 8051 Assembly
language
Show how the flag register is affected by the following instruction.
MOV A,#0F5H ;A=F5 hex
ADD A,#0BH ;A=F5+0B=00
Solution:
F5H 1111 0101
+ 0BH + 0000 1011
100H 0000 0000
CY =1, since there is a
carry out from D7
PF =1, because the number
of 1s is zero (an even
number), PF is set to 1.
AC =1, since there is a
carry from D3 to D4

HANEL
ARITHMETIC
INSTRUCTIONS
Addition of
Individual
Bytes
Assume that RAM locations 40 – 44H have the following values.
Write a program to find the sum of the values. At the end of the
program, register A should contain the low byte and R7 the high byte.
40 = (7D)
41 = (EB)
42 = (C5)
43 = (5B)
44 = (30)
Solution:
MOV R0,#40H ;load pointer
MOV R2,#5 ;load counter
CLR A ;A=0
MOV R7,A ;clear R7
AGAIN: ADD A,@R0 ;add the byte ptr to by R0
JNC NEXT ;if CY=0 don’t add carry
INC R7 ;keep track of carry
NEXT: INC R0 ;increment pointer
DJNZ R2,AGAIN ;repeat until R2 is zero

HANEL
ARITHMETIC
INSTRUCTIONS
ADDC and
Addition of 16-
Bit Numbers
When adding two 16-bit data operands,
the propagation of a carry from lower
byte to higher byte is concerned
Write a program to add two 16-bit numbers. Place the sum in R7 and
R6; R6 should have the lower byte.
Solution:
CLR C ;make CY=0
MOV A, #0E7H ;load the low byte now A=E7H
ADD A, #8DH ;add the low byte
MOV R6, A ;save the low byte sum in R6
MOV A, #3CH ;load the high byte
ADDC A, #3BH ;add with the carry
MOV R7, A ;save the high byte sum
When the first byte is added
(E7+8D=74, CY=1).
The carry is propagated to the
higher byte, which result in 3C
+ 3B + 1 =78 (all in hex)
1
3C E7
+ 3B 8D
78 74

HANEL
ARITHMETIC
INSTRUCTIONS
BCD Number
System
The binary representation of the digits
0 to 9 is called BCD (Binary Coded
Decimal) Digit BCD
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
8 1000
9 1001
Unpacked BCD
In unpacked BCD, the lower 4
bits of the number represent the
BCD number, and the rest of the
bits are 0
Ex. 00001001 and 00000101 are
unpacked BCD for 9 and 5
Packed BCD
In packed BCD, a single byte has
two BCD number in it, one in the
lower 4 bits, and one in the
upper 4 bits
Ex. 0101 1001 is packed BCD for
59H

HANEL
ARITHMETIC
INSTRUCTIONS
Unpacked and
Packed BCD
Adding two BCD numbers must give a
BCD result
MOV A, #17H
ADD A, #28H
The result above should have been 17 + 28 = 45 (0100 0101).
To correct this problem, the programmer must add 6 (0110) to the
low digit: 3F + 06 = 45H.
Adding these two
numbers gives
0011 1111B (3FH),
Which is not BCD!

HANEL
ARITHMETIC
INSTRUCTIONS
DA Instruction
DA A ;decimal adjust for addition
The DA instruction is provided to
correct the aforementioned problem
associated with BCD addition
The DA instruction will add 6 to the lower
nibble or higher nibble if need
Example:
MOV A,#47H ;A=47H first BCD operand
MOV B,#25H ;B=25H second BCD operand
ADD A,B ;hex(binary) addition(A=6CH)
DA A ;adjust for BCD addition
(A=72H)
The “DA” instruction works only on A. In other word, while the source
can be an operand of any addressing mode, the destination must be in
register A in order for DA to work.
DA works only
after an ADD,
but not after INC
6CH
72H

HANEL
ARITHMETIC
INSTRUCTIONS
DA Instruction
(cont’)
Summary of DA instruction
After an ADD or ADDC instruction
1. If the lower nibble (4 bits) is greater than 9, or
if AC=1, add 0110 to the lower 4 bits
2. If the upper nibble is greater than 9, or if
CY=1, add 0110 to the upper 4 bits
Example:
HEX BCD
29 0010 1001
+ 18 + 0001 1000
41 0100 0001 AC=1
+ 6 + 0110
47 0100 0111
Since AC=1 after the
addition, ”DA A” will add 6 to the
lower nibble.
The final result is in BCD format.

HANEL
ARITHMETIC
INSTRUCTIONS
DA Instruction
(cont’)
Assume that 5 BCD data items are stored in RAM locations starting
at 40H, as shown below. Write a program to find the sum of all the
numbers. The result must be in BCD.
40=(71)
41=(11)
42=(65)
43=(59)
44=(37)
Solution:
MOV R0,#40H ;Load pointer
MOV R2,#5 ;Load counter
CLR A ;A=0
MOV R7,A ;Clear R7
AGAIN: ADD A,@R0 ;add the byte pointer
;to by R0
DA A ;adjust for BCD
JNC NEXT ;if CY=0 don’t
;accumulate carry
INC R7 ;keep track of carries
NEXT: INC R0 ;increment pointer
DJNZ R2,AGAIN ;repeat until R2 is 0

HANEL
ARITHMETIC
INSTRUCTIONS
Subtraction of
Unsigned
Numbers
In many microprocessor there are two
different instructions for subtraction:
SUB and SUBB (subtract with borrow)
In the 8051 we have only SUBB
The 8051 uses adder circuitry to perform
the subtraction
SUBB A,source ;A = A – source – CY
To make SUB out of SUBB, we have to
make CY=0 prior to the execution of
the instruction
Notice that we use the CY flag for the
borrow

HANEL
ARITHMETIC
INSTRUCTIONS
Subtraction of
Unsigned
Numbers
(cont’)
SUBB when CY = 0
1. Take the 2’s complement of the
subtrahend (source operand)
2. Add it to the minuend (A)
3. Invert the carry
CLR C
MOV A,#4C ;load A with value 4CH
SUBB A,#6EH ;subtract 6E from A
JNC NEXT ;if CY=0 jump to NEXT
CPL A ;if CY=1, take 1’s complement
INC A ;and increment to get 2’s comp
NEXT: MOV R1,A ;save A in R1
Solution:
4C 0100 1100 0100 1100
- 6E 0110 1110 1001 0010
-22 01101 1110
CY =1
+
2’s
complement
Invert carry
CY=0, the result is positive;
CY=1, the result is negative
and the destination has the
2’s complement of the result

HANEL
ARITHMETIC
INSTRUCTIONS
Subtraction of
Unsigned
Numbers
(cont’)
SUBB when CY = 1
This instruction is used for multi-byte
numbers and will take care of the borrow
of the lower operand
CLR C
MOV A,#62H ;A=62H
SUBB A,#96H ;62H-96H=CCH with CY=1
MOV R7,A ;save the result
MOV A,#27H ;A=27H
SUBB A,#12H ;27H-12H-1=14H
MOV R6,A ;save the result
Solution:
We have 2762H - 1296H = 14CCH.
A = 62H – 96H – 0 = CCH
CY = 1
A = 27H - 12H - 1 = 14H
CY = 0

HANEL
ARITHMETIC
INSTRUCTIONS
Unsigned
Multiplication
The 8051 supports byte by byte
multiplication only
The byte are assumed to be unsigned data
MUL AB ;AxB, 16-bit result in B, A
MOV A,#25H ;load 25H to reg. A
MOV B,#65H ;load 65H to reg. B
MUL AB ;25H * 65H = E99 where
;B = OEH and A = 99H
Multiplication Operand1 Operand2 Result
Byte x byte A B B = high byte
A = low byte
Unsigned Multiplication Summary (MUL AB)

HANEL
ARITHMETIC
INSTRUCTIONS
Unsigned
Division
The 8051 supports byte over byte
division only
The byte are assumed to be unsigned data
DIV AB ;divide A by B, A/B
MOV A,#95 ;load 95 to reg. A
MOV B,#10 ;load 10 to reg. B
MUL AB ;A = 09(quotient) and
;B = 05(remainder)
Division Numerator Denominator Quotient Remainder
Byte / byte A B A B
Unsigned Division Summary (DIV AB)
CY is always 0
If B ≠ 0, OV = 0
If B = 0, OV = 1 indicates error

HANEL
ARITHMETIC
INSTRUCTIONS
Application for
DIV
(a) Write a program to get hex data in the range of 00 – FFH from
port 1 and convert it to decimal. Save it in R7, R6 and R5.
(b) Assuming that P1 has a value of FDH for data, analyze program.
Solution:
(a)
MOV A,#0FFH
MOV P1,A ;make P1 an input port
MOV A,P1 ;read data from P1
MOV B,#10 ;B=0A hex
DIV AB ;divide by 10
MOV R7,B ;save lower digit
MOV B,#10
DIV AB ;divide by 10 once more
MOV R6,B ;save the next digit
MOV R5,A ;save the last digit
(b) To convert a binary (hex) value to decimal, we divide it by 10
repeatedly until the quotient is less than 10. After each division the
remainder is saves.
Q R
FD/0A = 19 3 (low digit)
19/0A = 2 5 (middle digit)
2 (high digit)
Therefore, we have FDH=253.

HANEL
SIGNED
ARITHMETIC
INSTRUCTIONS
Signed 8-bit
Operands
D7 (MSB) is the sign and D0 to D6 are
the magnitude of the number
If D7=0, the operand is positive, and if
D7=1, it is negative
Positive numbers are 0 to +127
Negative number representation (2’s
complement)
1. Write the magnitude of the number in 8-bit
binary (no sign)
2. Invert each bit
3. Add 1 to it
Sign Magnitude
D7 D6 D5 D4 D3 D2 D1 D0

HANEL
SIGNED
ARITHMETIC
INSTRUCTIONS
Signed 8-bit
Operands
(cont’)
Show how the 8051 would represent -34H
Solution:
1. 0011 0100 34H given in binary
2. 1100 1011 invert each bit
3. 1100 1100 add 1 (which is CC in hex)
Signed number representation of -34 in 2’s complement is CCH
Decimal Binary Hex
-128 1000 0000 80
-127 1000 0001 81
-126 1000 0010 82
... ... ... ...
-2 1111 1110 FE
-1 1111 1111 FF
0 0000 0000 00
+1 0000 0001 01
+2 0000 0010 02
... ... ... ...
+127 0111 1111 7F

HANEL
SIGNED
ARITHMETIC
INSTRUCTIONS
Overflow
Problem
Examine the following code and analyze the result.
MOV A,#+96 ;A=0110 0000 (A=60H)
MOV R1,#+70 ;R1=0100 0110(R1=46H)
ADD A,R1 ;A=1010 0110
;A=A6H=-90,INVALID
Solution:
+96 0110 0000
+ +70 0100 0110
+ 166 1010 0110 and OV =1
According to the CPU, the result is -90, which is wrong. The CPU
sets OV=1 to indicate the overflow
If the result of an operation on signed
numbers is too large for the register
An overflow has occurred and the
programmer must be noticed

HANEL
SIGNED
ARITHMETIC
INSTRUCTIONS
OV Flag
In 8-bit signed number operations,
OV is set to 1 if either occurs:
1. There is a carry from D6 to D7, but no
carry out of D7 (CY=0)
2. There is a carry from D7 out (CY=1), but
no carry from D6 to D7
MOV A,#-128 ;A=1000 0000(A=80H)
MOV R4,#-2 ;R4=1111 1110(R4=FEH)
ADD A,R4 ;A=0111 1110(A=7EH=+126,INVALID)
-128 1000 0000
+ -2 1111 1110
-130 0111 1110 and OV=1
OV = 1
The result +126 is wrong

HANEL
SIGNED
ARITHMETIC
INSTRUCTIONS
OV Flag
(cont’)
MOV A,#-2 ;A=1111 1110(A=FEH)
MOV R1,#-5 ;R1=1111 1011(R1=FBH)
ADD A,R1 ;A=1111 1001(A=F9H=-7,
;Correct, OV=0)
-2 1111 1110
+ -5 1111 1011
-7 1111 1001 and OV=0
OV = 0
The result -7 is correct
MOV A,#+7 ;A=0000 0111(A=07H)
MOV R1,#+18 ;R1=0001 0010(R1=12H)
ADD A,R1 ;A=0001 1001(A=19H=+25,
;Correct,OV=0)
7 0000 0111
+ 18 0001 0010
25 0001 1001 and OV=0
OV = 0
The result +25 is correct

HANEL
SIGNED
ARITHMETIC
INSTRUCTIONS
OV Flag
(cont’)
In unsigned number addition, we must
monitor the status of CY (carry)
Use JNC or JC instructions
In signed number addition, the OV
(overflow) flag must be monitored by
the programmer
JB PSW.2 or JNB PSW.2

HANEL
SIGNED
ARITHMETIC
INSTRUCTIONS
2's
Complement
To make the 2’s complement of a
number
CPL A ;1’s complement (invert)
ADD A,#1 ;add 1 to make 2’s comp.

HANEL
LOGIC AND
COMPARE
INSTRUCTIONS
AND
ANL destination,source
;dest = dest AND source
This instruction will perform a logic
AND on the two operands and place
the result in the destination
The destination is normally the
accumulator
The source operand can be a register, in
memory, or immediate
X Y X AND Y
0 0 0
0 1 0
1 0 0
1 1 1
Show the results of the following.
MOV A,#35H ;A = 35H
ANL A,#0FH ;A = A AND 0FH
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
ANL is often used to
mask (set to 0) certain
bits of an operand

HANEL
LOGIC AND
COMPARE
INSTRUCTIONS
OR
ORL destination,source
;dest = dest OR source
The destination and source operands
are ORed and the result is placed in
the destination
accumulator
X Y X OR Y
0 0 0
0 1 1
1 0 1
1 1 1
MOV A,#04H ;A = 04
ORL A,#68H ;A = 6C
04H 0 0 0 0 0 1 0 0
68H 0 1 1 0 1 0 0 0
6CH 0 1 1 0 1 1 0 0
ORL instruction can be
used to set certain bits
of an operand to 1

HANEL
LOGIC AND
COMPARE
INSTRUCTIONS
XOR
XRL destination,source
;dest = dest XOR source
This instruction will perform XOR
operation on the two operands and
place the result in the destination
accumulator
MOV A,#54H
XRL A,#78H
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
X Y X XOR Y
0 0 0
0 1 1
1 0 1
1 1 0
XRL instruction can be
used to toggle certain
bits of an operand

HANEL
LOGIC AND
COMPARE
INSTRUCTIONS
XOR
(cont’)
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
AH = 45H.
45H 0 1 0 0 0 1 0 1
45H 0 1 0 0 0 1 0 1
00H 0 0 0 0 0 0 0 0
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 is not 0
MOV P2,#99H
EXIT: ...
XRL can be used to
see if two registers
have the same value
If both registers have the same
value, 00 is placed in A. JNZ
and JZ test the contents of the
accumulator.

HANEL
LOGIC AND
COMPARE
INSTRUCTIONS
Complement
Accumulator
CPL A ;complements the register A
This is called 1’s complement
To get the 2’s complement, all we
have to do is to to add 1 to the 1’s
complement
MOV A, #55H
CPL A ;now A=AAH
;0101 0101(55H)
;becomes 1010 1010(AAH)

HANEL
LOGIC AND
COMPARE
INSTRUCTIONS
Compare
Instruction
CJNE destination,source,rel. addr.
The actions of comparing and jumping
are combined into a single instruction
called CJNE (compare and jump if not
equal)
The CJNE instruction compares two
operands, and jumps if they are not equal
The destination operand can be in the
accumulator or in one of the Rn registers
The source operand can be in a register, in
The operands themselves remain unchanged
It changes the CY flag to indicate if the
destination operand is larger or smaller

HANEL
LOGIC AND
COMPARE
INSTRUCTIONS
Compare
Instruction
(cont’)
Notice in the CJNE instruction that any
Rn register can be compared with an
immediate value
There is no need for register A to be
involved
CJNE R5,#80,NOT_EQUAL ;check R5 for 80
... ;R5 = 80
NOT_EQUAL:
JNC NEXT ;jump if R5 > 80
... ;R5 < 80
NEXT: ...
Compare Carry Flag
destination ≥ source CY = 0
destination < source CY = 1
CY flag is always
checked for cases
of greater or less
than, but only after
it is determined that
they are not equal

HANEL
LOGIC AND
COMPARE
INSTRUCTIONS
Compare
Instruction
(cont’)
The compare instruction is really a
subtraction, except that the operands
remain unchanged
Flags are changed according to the
execution of the SUBB instruction
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 = 75 then A = 75
If T < 75 then R1 = T
If T > 75 then R2 = T
Solution:
MOV P1,#0FFH ;make P1 an input port
MOV A,P1 ;read P1 port
CJNE A,#75,OVER ;jump if A is not 75
SJMP EXIT ;A=75, exit
OVER: JNC NEXT ;if CY=0 then A>75
MOV R1,A ;CY=1, A<75, save in R1
SJMP EXIT ; and exit
NEXT: MOV R2,A ;A>75, save it in R2
EXIT: ...

HANEL
ROTATE
INSTRUCTION
AND DATA
SERIALIZATION
Rotating Right
and Left
RR A ;rotate right A
In rotate right
The 8 bits of the accumulator are rotated
right one bit, and
Bit D0 exits from the LSB and enters into
MSB, D7
MSB LSB
MOV A,#36H ;A = 0011 0110
RR A ;A = 0001 1011
RR A ;A = 1000 1101
RR A ;A = 1100 0110
RR A ;A = 0110 0011

HANEL
ROTATE
INSTRUCTION
AND DATA
SERIALIZATION
Rotating Right
and Left
(cont’)
RL A ;rotate left A
In rotate left
The 8 bits of the accumulator are rotated
left one bit, and
Bit D7 exits from the MSB and enters into
LSB, D0
MSB LSB
MOV A,#72H ;A = 0111 0010
RL A ;A = 1110 0100
RL A ;A = 1100 1001

HANEL
ROTATE
INSTRUCTION
AND DATA
SERIALIZATION
Rotating
through Carry
RRC A ;rotate right through carry
In RRC A
Bits are rotated from left to right
They exit the LSB to the carry flag, and
the carry flag enters the MSB
MSB LSB
CLR C ;make CY = 0
MOV A,#26H ;A = 0010 0110
RRC A ;A = 0001 0011 CY = 0
RRC A ;A = 0000 1001 CY = 1
RRC A ;A = 1000 0100 CY = 1
CY

HANEL
ROTATE
INSTRUCTION
AND DATA
SERIALIZATION
Rotating
through Carry
(cont’)
RLC A ;rotate left through carry
In RLC A
Bits are shifted from right to left
They exit the MSB and enter the carry flag,
and the carry flag enters the LSB
MSB LSBCY
Write a program that finds the number of 1s in a given byte.
MOV R1,#0
MOV R7,#8 ;count=08
MOV A,#97H
AGAIN: RLC A
JNC NEXT ;check for CY
INC R1 ;if CY=1 add to count
NEXT: DJNZ R7,AGAIN

HANEL
ROTATE
INSTRUCTION
AND DATA
SERIALIZATION
Serializing Data
Serializing data is a way of sending a
byte of data one bit at a time through
a single pin of microcontroller
Using the serial port, discussed in Chapter
10
To transfer data one bit at a time and
control the sequence of data and spaces
in between them

HANEL
ROTATE
INSTRUCTION
AND DATA
SERIALIZATION
Serializing Data
(cont’)
Transfer a byte of data serially by
Moving CY to any pin of ports P0 – P3
Using rotate instruction
Write a program to transfer value 41H serially (one bit at a time)
via pin P2.1. Put two highs at the start and end of the data. Send the
byte LSB first.
Solution:
MOV A,#41H
SETB P2.1 ;high
SETB P2.1 ;high
MOV R5,#8
AGAIN: RRC A
MOV P2.1,C ;send CY to P2.1
DJNZ R5,HERE
SETB P2.1 ;high
SETB P2.1 ;high
P2.1CYRegister A
D7 D0
Pin

HANEL
ROTATE
INSTRUCTION
AND DATA
SERIALIZATION
Serializing Data
(cont’)
Write a program to bring in a byte of data serially one bit at a time
via pin P2.7 and save it in register R2. The byte comes in with the
LSB first.
Solution:
MOV R5,#8
AGAIN: MOV C,P2.7 ;bring in bit
RRC A
DJNZ R5,HERE
MOV R2,A ;save it
P2.7 CY Register A
D7 D0
Pin

HANEL
ROTATE
INSTRUCTION
AND DATA
SERIALIZATION
Single-bit
Operations with
CY
There are several instructions by which
the CY flag can be manipulated directly
Instruction Function
SETB C Make CY = 1
CLR C Clear carry bit (CY = 0)
CPL C Complement carry bit
MOV b,C Copy carry status to bit location (CY = b)
MOV C,b Copy bit location status to carry (b = CY)
JNC target Jump to target if CY = 0
JC target Jump to target if CY = 1
ANL C,bit AND CY with bit and save it on CY
ANL C,/bit AND CY with inverted bit and save it on CY
ORL C,bit OR CY with bit and save it on CY
ORL C,/bit OR CY with inverted bit and save it on CY

HANEL
ROTATE
INSTRUCTION
AND DATA
SERIALIZATION
Single-bit
Operations with
CY
(cont’)
Assume that bit P2.2 is used to control an outdoor light and bit P2.5
a light inside a building. Show how to turn on the outside light and
turn off the inside one.
Solution:
SETB C ;CY = 1
ORL C,P2.2 ;CY = P2.2 ORed w/ CY
MOV P2.2,C ;turn it on if not on
CLR C ;CY = 0
ANL C,P2.5 ;CY = P2.5 ANDed w/ CY
MOV P2.5,C ;turn it off if not off
Write a program that finds the number of 1s in a given byte.
Solution:
MOV R1,#0 ;R1 keeps number of 1s
MOV R7,#8 ;counter, rotate 8 times
MOV A,#97H ;find number of 1s in 97H
AGAIN: RLC A ;rotate it thru CY
JNC NEXT ;check CY
INC R1 ;if CY=1, inc count
NEXT: DJNZ R7,AGAIN ;go thru 8 times

HANEL
ROTATE
INSTRUCTION
AND DATA
SERIALIZATION
SWAP
SWAP A
It swaps the lower nibble and the
higher nibble
In other words, the lower 4 bits are put
into the higher 4 bits and the higher 4 bits
are put into the lower 4 bits
SWAP works only on the accumulator
(A)
D7-D4before :
after : D3-D0
D3-D0
D7-D4

HANEL
ROTATE
INSTRUCTION
AND DATA
SERIALIZATION
SWAP
(cont’)
(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 = 0111 0010
RL A ;A = 0111 0010
RL A ;A = 0111 0010
RL A ;A = 0111 0010
RL A ;A = 0111 0010

HANEL
BCD AND ASCII
APPLICATION
PROGRAMS
Key ASCII (hex) Binary BCD (unpacked)
0 30 011 0000 0000 0000
1 31 011 0001 0000 0001
2 32 011 0010 0000 0010
3 33 011 0011 0000 0011
4 34 011 0100 0000 0100
5 35 011 0101 0000 0101
6 36 011 0110 0000 0110
7 37 011 0111 0000 0111
8 38 011 1000 0000 1000
9 39 011 1001 0000 1001
ASCII code and BCD for digits 0 - 9

HANEL
BCD AND ASCII
APPLICATION
PROGRAMS
Packed BCD to
ACSII
Conversion
The DS5000T microcontrollers have a
real-time clock (RTC)
The RTC provides the time of day (hour,
minute, second) and the date (year,
month, day) continuously, regardless of
whether the power is on or off
However this data is provided in
packed BCD
To be displayed on an LCD or printed by
the printer, it must be in ACSII format
Packed BCD Unpacked BCD ASCII
29H 02H & 09H 32H & 39H
0010 1001 0000 0010 & 0011 0010 &
0000 1001 0011 1001

HANEL
BCD AND ASCII
APPLICATION
PROGRAMS
ASCII to
Packed BCD
Conversion
To convert ASCII to packed BCD
It is first converted to unpacked BCD (to
get rid of the 3)
Combined to make packed BCD
key ASCII Unpacked BCD Packed BCD
4 34 0000 0100
7 37 0000 0111 0100 0111 or 47H
MOV A, #’4’ ;A=34H, hex for ‘4’
MOV R1,#’7’ ;R1=37H,hex for ‘7’
ANL A, #0FH ;mask upper nibble (A=04)
ANL R1,#0FH ;mask upper nibble (R1=07)
SWAP A ;A=40H
ORL A, R1 ;A=47H, packed BCD

HANEL
BCD AND ASCII
APPLICATION
PROGRAMS
ASCII to
Packed BCD
Conversion
(cont’)
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.
MOV A,#29H ;A=29H, packed BCD
MOV R2,A ;keep a copy of BCD data
ANL A,#0FH ;mask the upper nibble (A=09)
ORL A,#30H ;make it an ASCII, A=39H(‘9’)
MOV R6,A ;save it
MOV A,R2 ;A=29H, get the original
data
ANL A,#0F0H ;mask the lower nibble
RR A ;rotate right
RR A ;rotate right
RR A ;rotate right
RR A ;rotate right
ORL A,#30H ;A=32H, ASCII char. ’2’
MOV R2,A ;save ASCII char in R2
SWAP A

HANEL
BCD AND ASCII
APPLICATION
PROGRAMS
Using a Look-
up Table for
ASCII
Assume that the lower three bits of P1 are connected to three
switches. Write a program to send the following ASCII characters
to P2 based on the status of the switches.
000 ‘0’
001 ‘1’
010 ‘2’
011 ‘3’
100 ‘4’
101 ‘5’
110 ‘6’
111 ‘7’
Solution:
MOV DPTR,#MYTABLE
MOV A,P1 ;get SW status
ANL A,#07H ;mask all but lower 3
MOVC A,@A+DPTR ;get data from table
MOV P2,A ;display value
SJMP $ ;stay here
;------------------
ORG 400H
MYTABLE DB ‘0’,‘1’,‘2’,‘3’,‘4’,‘5’,‘6’,‘7’
END

HANEL
BCD AND ASCII
APPLICATION
PROGRAMS
Checksum Byte
in ROM
To ensure the integrity of the ROM
contents, every system must perform
the checksum calculation
The process of checksum will detect any
corruption of the contents of ROM
The checksum process uses what is called
a checksum byte
The checksum byte is an extra byte that is
tagged to the end of series of bytes of data

HANEL
BCD AND ASCII
APPLICATION
PROGRAMS
Checksum Byte
in ROM
(cont’)
To calculate the checksum byte of a
series of bytes of data
Add the bytes together and drop the
carries
Take the 2’s complement of the total sum,
and it becomes the last byte of the series
To perform the checksum operation,
add all the bytes, including the
checksum byte
The result must be zero
If it is not zero, one or more bytes of data
have been changed

HANEL
BCD AND ASCII
APPLICATION
PROGRAMS
Checksum Byte
in ROM
(cont’)
Assume that we have 4 bytes of hexadecimal data: 25H, 62H, 3FH, and
52H.(a) Find the checksum byte, (b) perform the checksum operation to
ensure data integrity, and (c) if the second byte 62H has been changed
to 22H, show how checksum detects the error.
Solution:
(a) Find the checksum byte.
25H The checksum is calculated by first adding the
+ 62H bytes. The sum is 118H, and dropping the carry,
+ 3FH we get 18H. The checksum byte is the 2’s
+ 52H complement of 18H, which is E8H
118H
(b) Perform the checksum operation to ensure data integrity.
25H
+ 62H Adding the series of bytes including the checksum
+ 3FH byte must result in zero. This indicates that all the
+ 52H bytes are unchanged and no byte is corrupted.
+ E8H
200H (dropping the carries)
(c) If the second byte 62H has been changed to 22H, show how
checksum detects the error.
25H
+ 22H Adding the series of bytes including the checksum
+ 3FH byte shows that the result is not zero, which indicates
+ 52H that one or more bytes have been corrupted.
+ E8H
1C0H (dropping the carry, we get C0H)

HANEL
BCD AND ASCII
APPLICATION
PROGRAMS
Binary (Hex)
to ASCII
Conversion
Many ADC (analog-to-digital converter)
chips provide output data in binary
(hex)
To display the data on an LCD or PC
screen, we need to convert it to ASCII
Convert 8-bit binary (hex) data to decimal
digits, 000 – 255
Convert the decimal digits to ASCII digits,
30H – 39H

6 arithmetic logic inst and prog

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