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:a 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
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