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
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
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.
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
