8085 Assembly Language Programs & Explanations
1.
Statement:
Store the data byte 32H into memory location 4000H.
Program 1:
MVI A, 32H :
Store 32H in the accumulator
STA 4000H :
Copy accumulator contents at address 4000H
HLT :
Terminate program execution
Program 2:
LXI H : Load
HL with 4000H
MVI M : Store
32H in memory location pointed by HL register pair
(4000H)
HLT :
Terminate program execution
2.
Statement: Exchange
the contents of memory locations 2000H and 4000H
Program 1:
LDA 2000H :
Get the contents of memory location 2000H into
accumulator
MOV B, A :
Save the contents into B register
LDA 4000H :
Get the contents of memory location 4000Hinto
accumulator
STA 2000H :
Store the contents of accumulator at address 2000H
MOV A, B : Get
the saved contents back into A register
STA 4000H :
Store the contents of accumulator at address 4000H
Program 2:
LXI H 2000H :
Initialize HL register pair as a pointer to
memory
location 2000H.
LXI D 4000H :
Initialize DE register pair as a pointer to
memory
location 4000H.
MOV B, M : Get
the contents of memory location 2000H into B
register.
LDAX D : Get
the contents of memory location 4000H into A
register.
MOV M, A :
Store the contents of A register into memory
location
2000H.
MOV A, B :
Copy the contents of B register into accumulator.
STAX D : Store
the contents of A register into memory location
4000H.
HLT :
Terminate program execution.
3.Sample
problem
(4000H) = 14H
(4001H) = 89H
Result = 14H +
89H = 9DH
Source program
LXI H 4000H :
HL points 4000H
MOV A, M : Get
first operand
INX H : HL
points 4001H
ADD M : Add
second operand
INX H : HL
points 4002H
MOV M, A :
Store result at 4002H
HLT :
Terminate program execution
4.Statement:
Subtract
the contents of memory location 4001H from the memory
location 2000H and place the
result in memory location 4002H.
Program - 4: Subtract two 8-bit numbers
Sample problem:
(4000H) = 51H
(4001H) = 19H
Result = 51H -
19H = 38H
Source
program:
LXI H, 4000H :
HL points 4000H
MOV A, M : Get
first operand
INX H : HL
points 4001H
SUB M :
Subtract second operand
INX H : HL
points 4002H
MOV M, A :
Store result at 4002H.
HLT :
Terminate program execution
5.Statement:
Add the
16-bit number in memory locations 4000H and 4001H to
the 16-bit number in memory
locations 4002H and 4003H. The most significant
eight bits of the two
numbers to be added are in memory locations 4001H and
4003H. Store the result in
memory locations 4004H and 4005H with the most
significant byte in memory
location 4005H.
Program - 5.a: Add two 16-bit numbers - Source Program 1
Sample problem:
(4000H) = 15H
(4001H) = 1CH
(4002H) = B7H
(4003H) = 5AH
Result = 1C15
+ 5AB7H = 76CCH
(4004H) = CCH
(4005H) = 76H
Source Program 1:
LHLD 4000H :
Get first I6-bit number in HL
XCHG : Save
first I6-bit number in DE
LHLD 4002H :
Get second I6-bit number in HL
MOV A, E : Get
lower byte of the first number
ADD L : Add
lower byte of the second number
MOV L, A :
Store result in L register
MOV A, D : Get
higher byte of the first number
ADC H : Add
higher byte of the second number with CARRY
MOV H, A : Store
result in H register
SHLD 4004H :
Store I6-bit result in memory locations 4004H and
4005H.
HLT :
Terminate program execution
6.Statement:
Add the
contents of memory locations 40001H and 4001H and place
the result in the memory
locations 4002Hand 4003H.
Sample problem:
(4000H) = 7FH
(400lH) = 89H
Result = 7FH +
89H = lO8H
(4002H) = 08H
(4003H) = 0lH
Source program:
LXI H, 4000H
:HL Points 4000H
MOV A, M :Get
first operand
INX H :HL
Points 4001H
ADD M :Add
second operand
INX H :HL
Points 4002H
MOV M, A :Store
the lower byte of result at 4002H
MVIA, 00
:Initialize higher byte result with 00H
ADC A :Add
carry in the high byte result
INX H :HL
Points 4003H
MOV M, A
:Store the higher byte of result at 4003H
HLT :Terminate
program execution
7.Statement:
Subtract
the 16-bit number in memory locations 4002H and 4003H
from the 16-bit number in
memory locations 4000H and 4001H. The most
significant eight bits of
the two numbers are in memory locations 4001H and 4003H.
Store the result in memory
locations 4004H and 4005H with the most significant
byte in memory location
4005H.
Sample problem
(4000H) = 19H
(400IH) = 6AH
(4004H) = I5H
(4003H) = 5CH
Result = 6A19H
- 5C15H = OE04H
(4004H) = 04H
(4005H) = OEH
Source program:
LHLD 4000H :
Get first 16-bit number in HL
XCHG : Save
first 16-bit number in DE
LHLD 4002H :
Get second 16-bit number in HL
MOV A, E : Get
lower byte of the first number
SUB L :
Subtract lower byte of the second number
MOV L, A :
Store the result in L register
MOV A, D : Get
higher byte of the first number
SBB H :
Subtract higher byte of second number with borrow
MOV H, A :
Store l6-bit result in memory locations 4004H and
4005H.
SHLD 4004H :
Store l6-bit result in memory locations 4004H and
4005H.
HLT :
Terminate program execution
8.Statement:
Find the
l's complement of the number stored at memory location
4400H and store the
complemented number at memory location 4300H.
Sample problem:
(4400H) = 55H
Result =
(4300B) = AAB
Source program:
LDA 4400B :
Get the number
CMA :
Complement number
STA 4300H : Store
the result
HLT :
Terminate program execution
9.Statement:
Find the
2's complement of the number stored at memory location
4200H and store the
complemented number at memory location 4300H.
Sample problem:
(4200H) = 55H
Result =
(4300H) = AAH + 1 = ABH
Source
program:
LDA 4200H :
Get the number
CMA :
Complement the number
ADI, 01 H :
Add one in the number
STA 4300H :
Store the result
HLT :
Terminate program execution
10.Statement:
Pack the
two unpacked BCD numbers stored in memory locations
4200H and 4201H and store
result in memory location 4300H. Assume the least
significant digit is stored
at 4200H.
Sample
problem:
(4200H) = 04
(4201H) = 09
Result =
(4300H) = 94
Source program
LDA 4201H :
Get the Most significant BCD digit
RLC
RLC
RLC
RLC : Adjust
the position of the second digit (09 is changed to
90)
ANI FOH : Make
least significant BCD digit zero
MOV C, A :
store the partial result
LDA 4200H :
Get the lower BCD digit
ADD C : Add
lower BCD digit
STA 4300H :
Store the result
HLT :
Terminate program execution
11.Statement:
Two
digit BCD number is stored in memory location 4200H.
Unpack the BCD number and
store the two digits in memory locations 4300H and
4301H such that memory
location 4300H will have lower BCD digit.
Sample problem
(4200H) = 58
Result = (4300H)
= 08 and
(4301H) = 05
Source program
LDA 4200H :
Get the packed BCD number
ANI FOH : Mask
lower nibble
RRC
RRC
RRC
RRC : Adjust
higher BCD digit as a lower digit
STA 4301H :
Store the partial result
LDA 4200H :
.Get the original BCD number
ANI OFH : Mask
higher nibble
STA 4201H :
Store the result
HLT :
Terminate program execution
12.Statement:Read the program given below
and state the contents of all
registers after the
execution of each instruction in sequence.
Main program:
4000H LXI SP,
27FFH
4003H LXI H,
2000H
4006H LXI B,
1020H
4009H CALL SUB
400CH HLT
Subroutine
program:
4100H SUB:
PUSH B
4101H PUSH H
4102H LXI B,
4080H
4105H LXI H,
4090H
4108H SHLD
2200H
4109H DAD B
410CH POP H
410DH POP B
410EH RET
13.Statement:Write a program to shift an
eight bit data four bits right. Assume
that data is in register C.
Source program:
MOV A, C
RAR
RAR
RAR
RAR
MOV C, A
HLT
14.Statement: Program to shift a 16-bit
data 1 bit left. Assume data is in the HL
register pair
Source program:
DAD H : Adds
HL data with HL data
15.Statement:
Write a
set of instructions to alter the contents of flag register in
8085.
PUSH PSW :
Save flags on stack
POP H :
Retrieve flags in 'L'
MOV A, L :
Flags in accumulator
CMA :
Complement accumulator
MOV L, A :
Accumulator in 'L'
PUSH H : Save
on stack
POP PSW : Back
to flag register
HLT :Terminate
program execution
16.Statement:
Calculate
the sum of series of numbers. The length of the series is
in memory location 4200H and
the series begins from memory location 4201H.
a. Consider the sum to be 8
bit number. So, ignore carries. Store the sum at memory
location 4300H.
b. Consider the sum to be 16
bit number. Store the sum at memory locations 4300H
and 4301H
a. Sample problem
4200H = 04H
4201H = 10H
4202H = 45H
4203H = 33H
4204H = 22H
Result = 10
+41 + 30 + 12 = H
4300H = H
Source program:
LDA 4200H
MOV C, A :
Initialize counter
SUB A : sum =
0
LXI H, 420lH :
Initialize pointer
BACK: ADD M :
SUM = SUM + data
INX H :
increment pointer
DCR C :
Decrement counter
JNZ BACK : if
counter 0 repeat
STA 4300H :
Store sum
HLT :
Terminate program execution
b. Sample problem
4200H = 04H
420lH = 9AH
4202H = 52H
4203H = 89H
4204H = 3EH
Result = 9AH +
52H + 89H + 3EH = H
4300H = B3H
Lower byte
4301H = 0lH
Higher byte
Source program:
LDA 4200H
MOV C, A : Initialize
counter
LXI H, 4201H :
Initialize pointer
SUB A :Sum low
= 0
MOV B, A : Sum
high = 0
BACK: ADD M :
Sum = sum + data
JNC SKIP
INR B : Add
carry to MSB of SUM
SKIP: INX H :
Increment pointer
DCR C :
Decrement counter
JNZ BACK :
Check if counter 0 repeat
STA 4300H :
Store lower byte
MOV A, B
STA 4301H :
Store higher byte
HLT :Terminate
program execution
17.Statement:
Multiply
two 8-bit numbers stored in memory locations 2200H and
2201H by repetitive addition
and store the result in memory locations 2300H and
2301H.
Sample problem:
(2200H) = 03H
(2201H) = B2H
Result = B2H +
B2H + B2H = 216H
= 216H
(2300H) = 16H
(2301H) = 02H
Source program
LDA 2200H
MOV E, A
MVI D, 00 :
Get the first number in DE register pair
LDA 2201H
MOV C, A :
Initialize counter
LX I H, 0000 H
: Result = 0
BACK: DAD D :
Result = result + first number
DCR C :
Decrement count
JNZ BACK : If
count 0 repeat
SHLD 2300H :
Store result
HLT :
Terminate program execution
18.Statement:Divide 16 bit number stored
in memory locations 2200H and 2201H
by the 8 bit number stored
at memory location 2202H. Store the quotient in memory
locations 2300H and 2301H
and remainder in memory locations 2302H and 2303H.
Sample problem
(2200H) = 60H
(2201H) = A0H
(2202H) = l2H
Result =
A060H/12H = 8E8H Quotient and 10H remainder
(2300H) = E8H
(2301H) = 08H
(2302H= 10H
(2303H) 00H
Source program
LHLD 2200H :
Get the dividend
LDA 2202H :
Get the divisor
MOV C, A
LXI D, 0000H :
Quotient = 0
BACK: MOV A, L
SUB C :
Subtract divisor
MOV L, A :
Save partial result
JNC SKIP : if
CY 1 jump
DCR H :
Subtract borrow of previous subtraction
SKIP: INX D :
Increment quotient
MOV A, H
CPI, 00 :
Check if dividend < divisor
JNZ BACK : if
no repeat
MOV A, L
CMP C
JNC BACK
SHLD 2302H :
Store the remainder
XCHG
SHLD 2300H :
Store the quotient
HLT :
Terminate program execution
19.Statement:Find the number of negative
elements (most significant bit 1) in a
block of data. The length of
the block is in memory location 2200H and the block
itself begins in memory
location 2201H. Store the number of negative elements in
memory location 2300H
Sample problem
(2200H) = 04H
(2201H) = 56H
(2202H) = A9H
(2203H) = 73H
(2204H) = 82H
Result = 02
since 2202H and 2204H contain numbers with a MSB of 1.
Source program
LDA 2200H
MOV C, A :
Initialize count
MVI B, 00 :
Negative number = 0
LXI H, 2201H :
Initialize pointer
BACK: MOV A, M
: Get the number
ANI 80H :
Check for MSB
JZ SKIP : If
MSB = 1
INR B :
Increment negative number count
SKIP: INX H :
Increment pointer
DCR C :
Decrement count
JNZ BACK : If
count 0 repeat
MOV A, B
STA 2300H :
Store the result
HLT :
Terminate program execution
20.Statement:Find the largest number in a
block of data. The length of the block
is in memory location 2200H
and the block itself starts from memory location
2201H.
Store the maximum number in
memory location 2300H. Assume that the numbers
in the block are all 8 bit
unsigned binary numbers.
Sample problem
(2200H) = 04
(2201H) = 34H
(2202H) = A9H
(2203H) = 78H
(2204H) =56H
Result =
(2202H) = A9H
Source program
LDA 2200H
MOV C, A :
Initialize counter
XRA A :
Maximum = Minimum possible value = 0
LXI H, 2201H :
Initialize pointer
BACK: CMP M :
Is number> maximum
JNC SKIP :
Yes, replace maximum
MOV A, M
SKIP: INX H
DCR C
JNZ BACK
STA 2300H :
Store maximum number
HLT :
Terminate program execution
21.Statement:Write a program to count
number of l's in the contents of D
register and store the count
in the B register.
Source program:
MVI B, 00H
MVI C, 08H
MOV A, D
BACK: RAR
JNC SKIP
INR B
SKIP: DCR C
JNZ BACK
HLT
22.Statement:Write a program to sort
given 10 numbers from memory location
2200H in the ascending
order.
Source program:
MVI B, 09 :
Initialize counter
START : LXI H,
2200H: Initialize memory pointer
MVI C, 09H :
Initialize counter 2
BACK: MOV A, M
: Get the number
INX H : Increment
memory pointer
CMP M :
Compare number with next number
JC SKIP : If
less, don't interchange
JZ SKIP : If
equal, don't interchange
MOV D, M
MOV M, A
DCX H
MOV M, D
INX H :
Interchange two numbers
SKIP:DCR C :
Decrement counter 2
JNZ BACK : If
not zero, repeat
DCR B :
Decrement counter 1
JNZ START
HLT :
Terminate program execution
23.Statement:Calculate the sum of series
of even numbers from the list of
numbers. The length of the
list is in memory location 2200H and the series itself
begins from memory location
2201H. Assume the sum to be 8 bit number so you can
ignore carries and store the
sum at memory location 2Sample
problem:
2200H= 4H
2201H= 20H
2202H= l5H
2203H= l3H
2204H= 22H
Result 22l0H=
20 + 22 = 42H
= 42H
Source program:
LDA 2200H
MOV C, A : Initialize
counter
MVI B, 00H :
sum = 0
LXI H, 2201H :
Initialize pointer
BACK: MOV A, M
: Get the number
ANI 0lH : Mask
Bit l to Bit7
JNZ SKIP :
Don't add if number is ODD
MOV A, B : Get
the sum
ADD M : SUM =
SUM + data
MOV B, A :
Store result in B register
SKIP: INX H :
increment pointer
DCR C :
Decrement counter
JNZ BACK : if
counter 0 repeat
STA 2210H :
store sum
HLT :
Terminate program execution
24.Statement:Calculate the sum of series
of odd numbers from the list of
numbers. The length of the
list is in memory location 2200H and the series itself
begins from memory location
2201H. Assume the sum to be 16-bit. Store the sum at
memory locations 2300H and
2301H.
Sample problem:
2200H = 4H
2201H= 9AH
2202H= 52H
2203H= 89H
2204H= 3FH
Result = 89H +
3FH = C8H
2300H= H Lower
byte
2301H = H
Higher byte
Source program
LDA 2200H
MOV C, A :
Initialize counter
LXI H, 2201H :
Initialize pointer
MVI E, 00 :
Sum low = 0
MOV D, E : Sum
high = 0
BACK: MOV A, M
: Get the number
ANI 0lH : Mask
Bit 1 to Bit7
JZ SKIP :
Don't add if number is even
MOV A, E : Get
the lower byte of sum
ADD M : Sum =
sum + data
MOV E, A :
Store result in E register
JNC SKIP
INR D : Add
carry to MSB of SUM
SKIP: INX H :
Increment pointer
DCR C :
Decrement
25.Statement:Find the square of the given
numbers from memory location 6100H
and store the result from
memory location 7000H
Source Program:
LXI H, 6200H :
Initialize lookup table pointer
LXI D, 6100H :
Initialize source memory pointer
LXI B, 7000H :
Initialize destination memory pointer
BACK: LDAX D :
Get the number
MOV L, A : A
point to the square
MOV A, M : Get
the square
STAX B : Store
the result at destination memory location
INX D :
Increment source memory pointer
INX B :
Increment destination memory pointer
MOV A, C
CPI 05H :
Check for last number
JNZ BACK : If
not repeat
HLT :
Terminate program execution
26.Statement:
Search
the given byte in the list of 50 numbers stored in the
consecutive memory locations
and store the address of memory location in the
memory locations 2200H and
2201H. Assume byte is in the C register and starting
address of the list is
2000H. If byte is not found store 00 at 2200H and 2201H.
Source program:
LX I H, 2000H
: Initialize memory pointer 52H
MVI B, 52H :
Initialize counter
BACK: MOV A, M
: Get the number
CMP C : Compare
with the given byte
JZ LAST : Go
last if match occurs
INX H :
Increment memory pointer
DCR B :
Decrement counter
JNZ B : I f
not zero, repeat
LXI H, 0000H
SHLD 2200H
JMP END :
Store 00 at 2200H and 2201H
LAST: SHLD
2200H : Store memory address
END: HLT :
Stop
27.Statement:
Two
decimal numbers six digits each, are stored in BCD package
form. Each number occupies a
sequence of byte in the memory. The starting
address of first number is
6000H Write an assembly language program that adds
these two numbers and stores
the sum in the same format starting from memory
location 6200H
Source Program:
LXI H, 6000H :
Initialize pointer l to first number
LXI D, 6l00H :
Initialize pointer2 to second number
LXI B, 6200H :
Initialize pointer3 to result
STC
CMC : Carry =
0
BACK: LDAX D :
Get the digit
ADD M : Add
two digits
DAA : Adjust
for decimal
STAX.B : Store
the result
INX H :
Increment pointer 1
INX D :
Increment pointer2
INX B :
Increment result pointer
MOV A, L
CPI 06H :
Check for last digit
JNZ BACK : If
not last digit repeat
HLT :
Terminate program execution
28.Statement:
Add 2
arrays having ten 8-bit numbers each and generate a third
array of result. It is
necessary to add the first element of array 1 with the first
element of array-2 and so
on. The starting addresses of array l, array2 and array3
are 2200H, 2300H and 2400H,
respectively.
Source Program:
LXI H, 2200H :
Initialize memory pointer 1
LXI B, 2300H :
Initialize memory pointer 2
LXI D, 2400H :
Initialize result pointer
BACK: LDAX B :
Get the number from array 2
ADD M : Add it
with number in array 1
STAX D : Store
the addition in array 3
INX H :
Increment pointer 1
INX B :
Increment pointer2
INX D :
Increment result pointer
MOV A, L
CPI 0AH :
Check pointer 1 for last number
JNZ BACK : If
not, repeat
HLT : Stop
29.Statement:
Write an
assembly language program to separate even numbers
from the given list of 50
numbers and store them in the another list starting from
2300H. Assume starting
address of 50 number list is 2200H
Source Program:
LXI H, 2200H :
Initialize memory pointer l
LXI D, 2300H :
Initialize memory pointer2
MVI C, 32H :
Initialize counter
BACK:MOV A, M
: Get the number
ANI 0lH :
Check for even number
JNZ SKIP : If
ODD, don't store
MOV A, M : Get
the number
STAX D : Store
the number in result list
INX D :
Increment pointer 2
SKIP: INX H :
Increment pointer l
DCR C :
Decrement counter
JNZ BACK : If
not zero, repeat
HLT : Stop
30.Statement:
Write
assembly language program with proper comments for the
following:
A block of data consisting
of 256 bytes is stored in memory starting at 3000H.
This block is to be shifted
(relocated) in memory from 3050H onwards. Do not shift
the block or part of the
block anywhere else in the memory.
Source Program:
Two blocks (3000 - 30FF and 3050 - 314F) are overlapping.
Therefore it
is necessary to transfer last byte first and first byte last.
MVI C, FFH :
Initialize counter
LX I H, 30FFH
: Initialize source memory pointer 3l4FH
LXI D, 314FH :
Initialize destination memory pointer
BACK: MOV A, M
: Get byte from source memory block
STAX D : Store
byte in the destination memory block
DCX H :
Decrement source memory pointer
DCX :
Decrement destination memory pointer
DCR C :
Decrement counter
JNZ BACK : If
counter 0 repeat
HLT : Stop
execution
31.Statement:
Add even
parity to a string of 7-bit ASCII characters. The length
of the string is in memory
location 2040H and the string itself begins in memory
location 2041H. Place even
parity in the most significant bit of each character.
Source Program:
LXI H, 2040H
MOV C ,M : Counter
for character
REPEAT:INX H :
Memory pointer to character
MOV A,M :
Character in accumulator
ORA A : ORing
with itself to check parity.
JPO PAREVEN :
If odd parity place
ORI 80H even
parity in D7 (80).
PAREVEN:MOV M
, A : Store converted even parity character.
DCR C :
Decrement counter.
JNZ REPEAT :
If not zero go for next character.
HLT
32.Statement:
A list
of 50 numbers is stored in memory, starting at 6000H. Find
number of negative, zero and
positive numbers from this list and store these results
in memory locations 7000H,
7001H, and 7002H respectively
Source Program:
LXI H, 6000H :
Initialize memory pointer
MVI C, 00H :
Initialize number counter
MVI B, 00H :
Initialize negative number counter
MVI E, 00H :
Initialize zero number counter
BEGIN:MOV A, M
: Get the number
CPI 00H : If
number = 0
JZ ZERONUM :
Goto zeronum
ANI 80H : If
MSB of number = 1i.e. if
JNZ NEGNUM
number is negative goto NEGNUM
INR D :
otherwise increment positive number counter
JMP LAST
ZERONUM:INR E
: Increment zero number counter
JMP LAST
NEGNUM:INR B :
Increment negative number counter
LAST:INX H :
Increment memory pointer
INR C :
Increment number counter
MOV A, C
CPI 32H : If
number counter = 5010 then
JNZ BEGIN :
Store otherwise check next number
LXI H, 7000 :
Initialize memory pointer.
MOV M, B :
Store negative number.
INX H
MOV M, E :
Store zero number.
INX H
MOV M, D :
Store positive number.
HLT :
Terminate execution
33.Statement:Write an 8085 assembly
language program to insert a string of four
characters from the tenth
location in the given array of 50 characters
Solution:
Step 1: Move bytes from location 10 till the end of array by
four bytes
downwards.
Step 2: Insert four bytes at locations 10, 11, 12 and 13.
Source Program:
LXI H, 2l31H :
Initialize pointer at the last location of array.
LXI D, 2l35H :
Initialize another pointer to point the last
location of
array after insertion.
AGAIN: MOV A,
M : Get the character
STAX D : Store
at the new location
DCX D :
Decrement destination pointer
DCX H :
Decrement source pointer
MOV A, L :
[check whether desired
CPI 05H bytes
are shifted or not]
JNZ AGAIN : if
not repeat the process
INX H : adjust
the memory pointer
LXI D, 2200H :
Initialize the memory pointer to point the string to
be inserted
REPE: LDAX D :
Get the character
MOV M, A :
Store it in the array
INX D :
Increment source pointer
INX H :
Increment destination pointer
MOV A, E :
[Check whether the 4 bytes
CPI 04 are
inserted]
JNZ REPE : if
not repeat the process
HLT : stop
34.Statement:Write an 8085 assembly
language program to delete a string of 4
characters from the tenth
location in the given array of 50 characters.
Solution: Shift bytes from location 14 till the end of array
upwards by 4
characters i.e. from location 10 onwards.
Source Program:
LXI H, 2l0DH
:Initialize source memory pointer at the 14thlocation
of the array.
LXI D, 2l09H :
Initialize destn memory pointer at the 10th location
of the array.
MOV A, M : Get
the character
STAX D : Store
character at new location
INX D :
Increment destination pointer
INX H : Increment
source pointer
MOV A, L :
[check whether desired
CPI 32H bytes
are shifted or not]
JNZ REPE : if
not repeat the process
HLT : stop
35.Statement:Multiply the 8-bit unsigned
number in memory location 2200H by
the 8-bit unsigned number in
memory location 2201H. Store the 8 least significant
bits of the result in memory
location 2300H and the 8 most significant bits in
memory location 2301H.
Sample problem:
(2200) = 1100
(0CH)
(2201) = 0101
(05H)
Multiplicand =
1100 (1210)
Multiplier =
0101 (510)
Result = 12 x
5 = (6010)
Source program
LXI H, 2200 :
Initialize the memory pointer
MOV E, M : Get
multiplicand
MVI D, 00H :
Extend to 16-bits
INX H :
Increment memory pointer
MOV A, M : Get
multiplier
LXI H, 0000 :
Product = 0
MVI B, 08H :
Initialize counter with count 8
MULT: DAD H :
Product = product x 2
RAL
JNC SKIP : Is
carry from multiplier 1 ?
DAD D : Yes,
Product =Product + Multiplicand
SKIP: DCR B :
Is counter = zero
JNZ MULT : no,
repeat
SHLD 2300H :
Store the result
HLT : End of
program
36.Statement:Divide the 16-bit unsigned
number in memory locations 2200H and
2201H (most significant bits
in 2201H) by the B-bit unsigned number in memory
location 2300H store the
quotient in memory location 2400H and remainder in
2401H
Assumption: The most significant bits of both the divisor and
dividend are
zero.
Source program
MVI E, 00 :
Quotient = 0
LHLD 2200H :
Get dividend
LDA 2300 : Get
divisor
MOV B, A :
Store divisor
MVI C, 08 :
Count = 8
NEXT: DAD H :
Dividend = Dividend x 2
MOV A, E
RLC
MOV E, A :
Quotient = Quotient x 2
MOV A, H
SUB B : Is
most significant byte of Dividend > divisor
JC SKIP : No,
go to Next step
MOV H, A :
Yes, subtract divisor
INR E : and
Quotient = Quotient + 1
SKIP:DCR C :
Count = Count - 1
JNZ NEXT : Is
count =0 repeat
MOV A, E
STA 2401H :
Store Quotient
Mov A, H
STA 2410H :
Store remainder
HLT : End of
program
37.DAA instruction is not
present. Write a sub routine which will perform the same
task as DAA.
Sample Problem:
Execution of DAA instruction:
1. If the value of the low order four bits (03-00) in the
accumulator is
greater than 9 or if auxiliary carry flag is set, the
instruction adds 6 '(06) to
the low-order four bits.
2. If the value of the high-order four bits (07-04) in the
accumulator is
greater than 9 or if carry flag is set, the instruction adds
6(06) to the highorder
four bits.
Source Program:
LXI SP, 27FFH
: Initialize stack pointer
MOV E, A :
Store the contents of accumulator
ANI 0FH : Mask
upper nibble
CPI 0A H :
Check if number is greater than 9
JC SKIP : if
no go to skip
MOV A, E : Get
the number
ADI 06H : Add
6 in the number
JMP SECOND :
Go for second check
SKIP: PUSH PSW
: Store accumulator and flag contents in stack
POP B : Get
the contents of accumulator in B register and
flag register
contents in C register
MOV A, C : Get
flag register contents in accumulator
ANI 10H :
Check for bit 4
JZ SECOND : if
zero, go for second check
MOV A, E : Get
the number
ADI 06 : Add 6
in the number
SECOND: MOV E,
A : Store the contents of accumulator
ANI FOH : Mask
lower nibble
RRC
RRC
RRC
RRC : Rotate
number 4 bit right
CPI 0AH :
Check if number is greater than 9
JC SKIPl : if
no go to skip 1
MOV A, E : Get
the number
ADI 60 H : Add
60 H in the number
JMP LAST : Go
to last
SKIP1: JNC
LAST : if carry flag = 0 go to last
MOV A, E : Get
the number
ADI 60 H : Add
60 H in the number
LAST: HLT
38.tement:To test RAM by writing '1'
and reading it back and later writing '0'
(zero) and reading it back.
RAM addresses to be checked are 40FFH to 40FFH. In
case of any error, it is
indicated by writing 01H at port 10H
Source Program:
LXI H, 4000H :
Initialize memory pointer
BACK: MVI M,
FFH : Writing '1' into RAM
MOV A, M :
Reading data from RAM
CPI FFH :
Check for ERROR
JNZ ERROR : If
yes go to ERROR
INX H :
Increment memory pointer
MOV A, H
CPI SOH : Check
for last check
JNZ BACK : If
not last, repeat
LXI H, 4000H :
Initialize memory pointer
BACKl: MVI M,
OOH : Writing '0' into RAM
MOV A, M :
Reading data from RAM
CPI OOH :
Check for ERROR
INX H :
Increment memory pointer
MOV A, H
CPI SOH :
Check for last check
JNZ BACKl : If
not last, repeat
HLT : Stop
Execution
39.Statement:Write an assembly language
program to generate fibonacci number
Source Program:
MVI D, COUNT : Initialize counter
MVI B, 00 :
Initialize variable to store previous number
MVI C, 01 : Initialize
variable to store current number
MOV A, B :[Add
two numbers]
BACK: ADD C
:[Add two numbers]
MOV B, C :
Current number is now previous number
MOV C, A :
Save result as a new current number
DCR D :
Decrement count
JNZ BACK : if
count 0 go to BACK
HLT : Stop
40.Statement:Write a program to generate
a delay of 0.4 sec if the crystal frequency
is 5 MHz
Calculation: In 8085, the operating frequency is half of the
crystal
frequency,
ie.Operating frequency = 5/2 = 2.5 MHz
Time for one T -state =
Number of T-states required =
= 1 x 106
Source Program:
LXI B, count :
16 - bit count
BACK: DCX B :
Decrement count
MOV A, C
ORA B :
Logically OR Band C
JNZ BACK : If
result is not zero repeat
41.Statement:
Arrange
an array of 8 bit unsigned no in descending order
Source Program:
START:MVI B,
00 ; Flag = 0
LXI H, 4150 ;
Count = length of array
MOV C, M
DCR C ; No. of
pair = count -1
INX H ; Point
to start of array
LOOP:MOV A, M ; Get kth element
INX H
CMP M ;
Compare to (K+1) th element
JNC LOOP 1 ;
No interchange if kth >= (k+1) th
MOV D, M ;
Interchange if out of order
MOV M, A ;
DCR H
MOV M, D
INX H
MVI B, 01H ;
Flag=1
LOOP 1:DCR C ;
count down
JNZ LOOP ;
DCR B ; is
flag = 1?
JZ START ; do
another sort, if yes
HLT ; If flag
= 0, step execution
42.Statement:
Transfer
ten bytes of data from one memory to another memory block.
Source memory block starts
from memory location 2200H where as destination
memory block starts from
memory location 2300H
Source Program:
LXI H, 4150 :
Initialize memory pointer
MVI B, 08 :
count for 8-bit
MVI A, 54
LOOP : RRC
JC LOOP1
MVI M, 00 :
store zero it no carry
JMP COMMON
LOOP2: MVI M,
01 : store one if there is a carry
COMMON: INX H
DCR B : check
for carry
JNZ LOOP
HLT :
Terminate the program
43.Statement:
Program
to calculate the factorial of a number between 0 to 8
Source program
LXI SP, 27FFH
; Initialize stack pointer
LDA 2200H ;
Get the number
CPI 02H ;
Check if number is greater than 1
JC LAST
MVI D, 00H ;
Load number as a result
MOV E, A
DCR A
MOV C,A ; Load
counter one less than number
CALL FACTO ;
Call subroutine FACTO
XCHG ; Get the
result in HL
SHLD 2201H ;
Store result in the memory
JMP END
LAST: LXI H,
000lH ; Store result = 01
END: SHLD
2201H
HLT
44.Statement:Write a program to find the
Square Root of an 8 bit binary number.
The binary number is stored
in memory location 4200H and store the square root in
4201H.
Source Program:
LDA 4200H :
Get the given data(Y) in A register
MOV B,A : Save
the data in B register
MVI C,02H :
Call the divisor(02H) in C register
CALL DIV :
Call division subroutine to get initial value(X)
in D-reg
REP: MOV E,D :
Save the initial value in E-reg
MOV A,B : Get
the dividend(Y) in A-reg
MOV C,D : Get
the divisor(X) in C-reg
CALL DIV :
Call division subroutine to get initial
value(Y/X) in
D-reg
MOV A, D :
Move Y/X in A-reg
ADD E : Get
the((Y/X) + X) in A-reg
MVI C, 02H :
Get the divisor(02H) in C-reg
CALL DIV :
Call division subroutine to get ((Y/X) + X)/2
in D-reg.This
is XNEW
MOV A, E : Get
Xin A-reg
CMP D :
Compare X and XNEW
JNZ REP : If
XNEW is not equal to X, then repeat
STA 4201H :
Save the square root in memory
HLT :
Terminate program execution
45.Statement:Write a simple program to
Split a HEX data into two nibbles and store
it in memory
Source Program:
LXI H, 4200H :
Set pointer data for array
MOV B,M : Get
the data in B-reg
MOV A,B : Copy
the data to A-reg
ANI OFH : Mask
the upper nibble
INX H :
Increment address as 4201
MOV M,A :
Store the lower nibble in memory
MOV A,B : Get
the data in A-reg
ANI FOH :
Bring the upper nibble to lower nibble position
RRC
RRC
RRC
RRC
INX H
MOV M,A :
Store the upper nibble in memory
HLT :
Terminate program execution
46.Statement:
Add two
4 digit BCD numbers in HL and DE register pairs and store
result in memory locations,
2300H and 2301H. Ignore carry after 16 bit.
Sample Problem:
(HL) =3629
(DE) =4738
Step 1 : 29 +
38 = 61 and auxiliary carry flag = 1
:.add 06
61 + 06 = 67
Step 2 : 36 +
47 + 0 (carry of LSB) = 7D
Lower nibble
of addition is greater than 9, so add 6.
7D + 06 = 83
Result = 8367
Source program
MOV A, L : Get
lower 2 digits of no. 1
ADD E : Add
two lower digits
DAA : Adjust
result to valid BCD
STA 2300H :
Store partial result
MOV A, H : Get
most significant 2 digits of number
ADC D : Add
two most significant digits
DAA : Adjust
result to valid BCD
STA 2301H :
Store partial result
HLT :
Terminate program execution
47.Statement:
Subtract
the BCD number stored in E register from the number stored
in the D register.
Source Program:
MVI A,99H
SUB E : Find
the 99's complement of subtrahend
INR A : Find
100's complement of subtrahend
ADD D : Add
minuend to 100's complement of subtrahend
DAA : Adjust
for BCD
HLT :
Terminate program execution
48.Statement:
Write an
assembly language program to multiply 2 BCD numbers
Source Program:
MVI C,
Multiplier : Load BCD multiplier
MVI B, 00 :
Initialize counter
LXI H, 0000H :
Result = 0000
MVI E,
multiplicand : Load multiplicand
MVI D, 00H :
Extend to 16-bits
BACK: DAD D :
Result Result + Multiplicand
MOV A, L : Get
the lower byte of the result
ADI, 00H
DAA : Adjust
the lower byte of result to BCD.
MOV L, A :
Store the lower byte of result
MOV A, H : Get
the higher byte of the result
ACI, 00H
DAA : Adjust
the higher byte of the result to BCD
MOV H, A :
Store the higher byte of result.
MOV A, B :
[Increment
ADI 01H :
counter
DAA : adjust
it to BCD and
MOV B,A :
store it]
CMP C :
Compare if count = multiplier
JNZ BACK : if
not equal repeat
HLT : Stop
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