Store 8-bit data in memory


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"
Note: 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.

Exchange the contents of memory locations


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."
Note: In Program 1, direct addressing instructions are used, whereas in Program 2, indirect addressing instructions are used.

Add two 8-bit numbers


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"


Subtract two 8-bit numbers


Statement: Subtract the contents of memory location 4001H from the memory location 2000H and place the result in memory location 4002H.
Program –: 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"




Add two 16-bit numbers


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"
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.

Add contents of two memory locations


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"


Subtract two 16-bit numbers


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"




Finding one’s complement of a number


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"


Finding Two’s complement of a number


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"


Pack the unpacked BCD numbers


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"


Unpack a BCD number


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"


Execution format of instructions


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


Right shift, bit of data( 8 bit and 16 bit)


Statement: Write a program to shift an eight bit data four bits right. Assume data is in register C.
Sample problem:
(4200H) = 58
Result = (4300H) = 08 and
    (4301H) = 05

Source program 1: 
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 2
MOV A, B
RAR
MOV B, A
MOV A, C
RAR
MOV C, A
HLT


Left Shifting of a 16-bit data


Statement: Program to shift a 16-bit data 1 bit left. Assume data is in the HL register
HL = 1025 = 0001 0000 0010 0101

    HL = 0001 0000 0010 0101
  + HL = 0001 0000 0010 0101
----------------------------
Result = 0010 0000 0100 1010                  


Alter the contents of flag register in 8085


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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