Final answer:
The subject is to write an 8085 microprocessor assembly language program for adding two 16-bit numbers. The process involves handling both the lower and higher 8 bits separately and considering the carry generated from the lower byte addition. A sample code shows how this can be successfully implemented.
Step-by-step explanation:
To write an 8085 microprocessor assembly language program to add two 16-bit numbers, we need to follow a step-by-step approach. Since the 8085 microprocessor is an 8-bit processor, we cannot add 16-bit numbers directly. Instead, we have to add them in two parts: the lower 8 bits (least significant byte) and the higher 8 bits (most significant byte), taking care of any carry that might arise from the addition of the lower bytes.
Here is a sample assembly language program that can perform this addition:
LXI H, 4000H ; HL points to the first number
MOV A, M ; Load lower byte of the first number into A
INX H
MOV B, M ; Load higher byte of the first number into B
LXI H, 4002H ; HL points to the second number
ADD M ; Add lower byte of the second number
JNC SKIP ; If no carry, jump to SKIP
INR B ; If carry, increment higher byte
SKIP: INX H
ADD B ; Add higher byte of the second number
STA 4004H ; Store result's lower byte at 4004H
INX H
STA 4005H ; Store result's higher byte at 4005H
The program assumes that the first 16-bit number is stored at memory addresses 4000H and 4001H, and the second 16-bit number is stored at 4002H and 4003H. The result is stored at 4004H (lower byte) and 4005H (higher byte).