| 8085.zip |
PROGRAMS/ALP FOR 8085
1. Statement: Add the contents of memory locations 4000H and 4001H and place the result in memory location 4002H.
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
2. Statement: Subtract the contents of memory location 4001H from the memory location 2000H and place the result in memory location 4002H.
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
3. 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.
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
4. 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.
5. 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
6. 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
7. 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
8. 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
9. 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
10.Statement:Write a program to sort given 10 numbers from memory location 2200H in the ascending order.
MVI B, 09 : Initialize counter
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
11.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 2210H.
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
12.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
13.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
14.Statement:Write an assembly language program to generate fibonacci number.
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.
15.Statement: Arrange an array of 8 bit unsigned no in descending order
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
16.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
Subroutine:
DIV: MVI D, 00H : Clear D-reg for Quotient
NEXT:SUB C : Subtact the divisor from dividend
INR D : Increment the quotient
CMP C : Repeat subtraction until the
JNC NEXT : divisor is less than dividend
RET : Return to main program
