Pack the two unpacked BCD numbers stored in memory locations 4200H and 4201H and store result in memory location 4300H. Assume the least significant d

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

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

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

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

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.

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 ei

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

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

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: 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.