The document discusses various logical and arithmetic operations in 8051 assembly language. It describes different types of instructions including data transfer, program branching, logical and arithmetic. Logical instructions include byte-level and bit-level logical operations like AND, OR, XOR. Arithmetic instructions cover operations like addition, subtraction, multiplication and division. Examples are provided to illustrate how these instructions work and how they affect the flag register.

1. Presented by
HEMANTH
12MT06PED011
ARITHMETIC
AND
LOGICAL
OPERATIONS

2. INSTRUCTION TYPES:
The 8051 instructions are divided into
1. Data transfer
2. Program branching
3. Logical
4. Arithmetic

3. LOGICAL INSTRUCTIONS:
Introduction
 Application in machine control
 Byte operators: for manipulating all 8051 RAM
 Bit operators: For bit addressable internal RAM area some
SFRs
 Bit operators yield compact program code and enhances
execution speed
TYPES:
1. Byte-Level logical operation
2. Bit-Level logical operations
3. Rotate and Swap operations

4. Bit and Byte level Instructions:
Boolean operator
• AND
• OR
• XOR
• NOT
8051 Mnemonic
• ANL(AND logical)
• ORL(OR logical)
• XRL(exclusive OR
logical)
• CPL(complement)
Byte Level Operations:
Entire bits of destination are effected.
Destination: register A or direct address in internal RAM
No flags are affected unless the direct address is PSW
Only internal RAM or SFRs may be logically manipulated

5. BYTE LEVEL INSTRUCTIONS

6. CPL A is called 1’s complement

7. EXAMPLE OF LOGIC OPERATIONS
MOV A, #OFFh ;A= FFh =1111 1111
MOV R0, #77h ;R0=77h =0111 0111
ANL A,R0 ;A =77h =0111 0111
MOV 15h,A ;15h contains 77h
CPL A ;A=88h=1000 1000
ORL 15h,#88h ;15h Contains FFh
XRL A,15h ;A=77h
XAL A,R0 ;A=00h
ANL A,15h ;A=00h
ORL A,R0 ;A=77h
CLR A ;A=00h
XRL 15h,A ;15h= FFh
XRL A,R0 ;A=77h

8. Bit Level Operation:
• All I/O ports and registers A, B, PSW, IP,
IE,SCON, and TCON are bit-addressable
• Bit-addressable SFRs:
• No flags, other than C flag, are affected, unless the
flag bit is an addressed bit.

9. Bit-level Boolean operations:
If the destination bit is a port bit, the SFR latch bit is
affected, not the pin.
ANL C,/b and ORL C,/b do not alter the addressed bit b.

10. Example: Write a program to save the accumulator in
R7 of bank 2.
Solution:
CLR PSW.3
SETB PSW.4 ;RS1=1& RS0=0 ;BANK 2 IS SELECTED
MOV R7,A

11. ROTATE & SWAP INSTRUCTIONS:
 Only The register A can be rotated one bit position to the
left or right with or without including the C flag
RRC and RLC affects only carry flag.
RL rotate a byte to left; MSB becomes LSB
RLC Rotate a byte and the carry bit left; the carry
bit becomes LSB, MSB becomes the carry.
RR Rotate a byte to right; LSB becomes MSB
RRC Rotate a byte and the carry bit right; LSB
becomes the carry, the carry the MSB
SWAP Exchange the low and high nibbles in a byte

12. RR A: Rotate right:
MOV A , #36H ;A= 0011 0110
RR A ; 0011 0110 ,so A= 0001 1011
RR A ;A=1000 1101
RL A: Rotate Left
MOV A,#72H ;A =0111 0010
RL A ; 0111 0010 ,so A = 1110 0100
RL A ;A = 1100 1001

13. RRC A: Rotate Right through Carry
CLR C ;CY = 0 MSB-LSB C
MOV A,#A5H ;A = 1010 0101
RRC A ;A = 0101 0010;CY = 1
RRC A ;A = 1010 1001;CY = 0
RLC A: Rotate Left through Carry
CLR C ;CY=0
MOV A,#6A ;A=0110 1010
RLC A ; A=1101 0100;CY=0
RLC A ;A=1010 1000;CY=1
MSB-LSB C

14. SWAP A
 It swaps the lower nibble and the higher nibble
 SWAP works only on the accumulator(A)
 E.g. MOV A,#72H ;A = 72H
SWAP A ;A = 27H

15. Write a program that finds the number of 1’s in a given
byte(097h).
MOV R1,#00
MOV R7,#08 ;count=08
MOV A,#097H
CLR C
AGAIN: RLC A
JNC NEXT ;check for CY
INC R1 ;if CY=1 add to count
NEXT: DJNZ R7,AGAIN

16. 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 ‘OR’ed with CY
MOV P2.2,C ;turn it on if not on
CLR C ;CY = 0
ANL C,P2.5 ;CY = P2.5 ‘AND’ed with CY
MOV P2.5,C ;turn it off if not off

17. ARITHMETIC OPERATIONS:
MNEMONIC OPERATION
INC destination Increment destination by 1
DEC destination Decrement destination by 1
ADD/ADDC destination, source Add source to destination
without/with carry( C ) flag
SUBB destination, source Subtract, with carry,
source from destination
MUL AB Multiply the contents of
registers A & B
DIV AB Divide the contents of register A by
contents of register B ( A/B )
DA A Decimal adjust the register A

19. INCREMENTING &
DECREMENTING

21. ADD
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

22. FLAGS AFFECTED:
C,AC,OV
• C flag is set to 1 if there is a carry out of bit
position 7; it is cleared to 0 otherwise.
• AC flag is set to 1 if there is a carry out of bit
position 3; it is cleared to 0 otherwise.
• OV flag is set to 1 if there is a carry out of bit
position 7,not bit position6 or if there is a carry out
of bit position6 not bit position7.

23. 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

24. SIGNED & UNSIGNED ADDITIION
• Unsigned numbers:8 bit magnitude
• Signed numbers: use bit 7 as a sign bit
 Bit 0-6 magnitude of no.
 Bit 7=1 means no. is negative
=0 : positive
In signed form, a single byte number range:-128d (1000
0000) to +127d(0111 1111)
In unsigned form:00h(0000 0000) to FFh(1111 1111)
• Adding or subtracting unsigned numbers may create
CY flag when exceeds FFh or a borrow when minuend
is less than subtrahend.
• In the case of signed numbers along with CY,OV is
used for sign bit actions.

25. • Unsigned addition:
There is a Carry out from sum, so
Carry flag is set to 1
Signed addition:. addition of Unlike signed numbers
Here there is a carry from bit 7,so CY=1. Also a carry from
bit 6 and OV=0. So no action is needed to correct the sum

26. Addition of like signed numbers:
• +100d(0110 0100=64h)
+050d(0011 0010=32h)
= +150d(1001 0110=96h) CY=0,OV=1

27. • -030d(1110 0010, 2’s cmpl of 0001 1110)+
-050d(1100 1110, 2’s cmpl of 0011 0010)
=-080d(1011 0000 ) CY=1.OV=1
SIGNED NO. 1011 0000=-48d,so we have to take
2’s complement of this answer.

28. ADDC: Add with carry:
ADDC IS NORMALLY USED TO ADD A CARRY AFTER
THE LSB ADDITION IN A MULTI BYTE PROCESS.

29. • ADD A,R5
Example:
1Ch= 0001 1100+
A1h= 1010 0001
BDh=1011 1101
• ADDC A,#10h:
• 5Eh=0101 1110+
10h=0001 0000=(6Eh=0110 1110)
6Eh+1(CY)=6Fh

30. SUBTRACTION:SUBB

31. MULTIPLICATION:MUL AB
• MUL AB ;multiply A by B; put the lower-order byte
of product in A and higher order byte in B
• Only registers A and B can be used.
• Use A and B as both as source and destination
address for operation
• Both numbers are unsigned
• OV will be set, If A*B>FFh, it means the product is
>8bit and Reg B should be checked for higher byte.
• CY is always cleared to 0.

32. Example :Multiply FFh with FFh
• A=FFh & B=FFh will give largest possible
product(FE01h).
• 01h will be stored in A and FEh in B.
• OV=1;CY=0
Example :

33. DIV AB :DIVISION
• DIV AB ;divide A by B; put the integer part of
quotient in Reg A and the integer part of the reminder in
B
• Only registers A and B can be used.
• Use A and B as both as source and destination address
for operation
• Both numbers are unsigned

34. Example:
*divide FFh by 2Ch:
255/44=5.8d
35d=23h

36. Byte cycle