Saturday, September 26, 2009
8085 programming (part1)
Statement: Store the data byte 32H into memory location 4000H.
Program 1:
MVI A, 52H : 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
The result of both programs will be the same. In program 1 direct addressing instruction is used, whereas in program 2 indirect addressing instruction is used.
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.
In Program 1, direct addressing instructions are used, whereas in Program 2, indirect addressing instructions are used.
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
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
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
Program - 5b: Add two 16-bit numbers - Source Program 2
Source program 2:
LHLD 4000H : Get first I6-bit number
XCHG : Save first I6-bit number in DE
LHLD 4002H : Get second I6-bit number in HL
DAD D : Add DE and HL
SHLD 4004H : Store I6-bit result in memory locations 4004H and 4005H.
HLT : Terminate program execution
NOTE: In program 1, eight bit addition instructions are used (ADD and ADC) and addition is performed in two steps. First lower byte addition using ADD instruction and then higher byte addition using ADC instruction.In program 2, 16-bit addition instruction (DAD) is used.
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
MVI A, 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
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
MVI A, 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
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
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, 01H : Add one in the number
STA 4300H : Store the result
HLT : Terminate program execution
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
NOTE:
BCD NO.: The numbers "0 to 9" are called BCD (Binary Coded Decimal) numbers. A decimal number 29 can be converted into BCD number by splitting it into two. ie. 02 and 09.
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
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
Note:
The table given gives the instruction sequence and the contents of all registers and stack after execution of each instruction.
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
Statement: Write a program to shift a 16 bit data, 1 bit right. Assume that data is in BC register pair.
Source program:
MOV A, B
RAR
MOV B, A
MOV A, C
RAR
MOV C, A
HLT
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
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
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
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
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
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
- Find the largest of given numbers
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
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
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
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.
Sample 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 Bit1 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
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 counter
JNZ BACK : Check if counter 0 repeat
MOV A, E
STA 2300H : Store lower byte
MOV A, D
STA 2301H : Store higher byte
HLT : Terminate program execution
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
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
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
- Add each element of array with the elements of another array
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
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
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
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 : Terminate program execution
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
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
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
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
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.
Statement:Assume the 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 high-order 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
Note: To check auxiliary carry flag it is necessary to get the flag register contents in one of the registers and then we can check the auxiliary carry flag by checking bit 4 of that register. To get the flag register contents in any general purpose register we require stack operation and therefore stack pointer is initialized at the beginning of the source program.
Statement: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
Labels: 8085, assembly language, microprocessors
DCX D
: September 3, 2017 at 2:36 AM 2-You can download the program in the dialer with the starting address of F200H. Compile it to obtain the output file. Using a simulator program analysis in table form on the affected registers and memory eyes. Do it by showing the step-by-step work. Total work of the program Find the time with the simulator and write it.
CAN ANYBODY HELP ME WİTH THİS
24000H by the data of memory location 24004H and then store
the quotient of the result at the memory location 2400BH and
the reminder of the result at 2400F. Finally clear the content of
24000H, 24004H.
please help me by using mov only
hawar_khalil@live.com
: January 23, 2021 at 1:14 PM Subscribe to Post Comments [Atom]
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