Q22Microprocessor and Microcontroller

When adding two 8-bit numbers, the maximum possible result ($FFH+FFH=1FEH$) requires 9 bits to mathematically represent. Therefore, an 8-bit microprocessor will overflow and the 9th bit is captured entirely by the Carry Flag (CY). A robust program must check this flag and store the carry as a separate byte to yield a correct 16-bit result.

Memory Architecture

• Input Data 1: Memory Address $2501H$
• Input Data 2: Memory Address $2502H$
• Output Sum (Lower 8-bits): Memory Address $2503H$
• Output Carry (Upper 8-bits): Memory Address $2504H$

Assembly Code Implementation

``assembly MVI C, 00H ; Initialize Register C to strictly 00H. This will explicitly hold our Carry. LXI H, 2501H ; Load the H-L register pair with the memory address of the first number. MOV A, M ; Fetch Data 1 from RAM (pointed to by HL) directly into the Accumulator. INX H ; Increment the H-L pair to point to 2502H (location of Data 2). ADD M ; Mathematically ADD Data 2 (from RAM) to the Accumulator. ; The sum is now in A. The Carry Flag (CY) is modified. JNC SKIP ; (Jump if No Carry). If CY=0, skip the next instruction. INR C ; If CY=1, increment Register C from 00H to 01H to record the carry. SKIP: INX H ; Increment H-L to point to 2503H (destination for the Sum). MOV M, A ; Store the final 8-bit Sum from the Accumulator directly into RAM. INX H ; Increment H-L to point to 2504H (destination for the Carry). MOV A, C ; Move the Carry value (00H or 01H) from Register C into Accumulator. MOV M, A ; Store the Carry value directly into RAM. HLT ; Halt the microprocessor execution. ``

Algorithmic Trace

Assume $2501H$ holds C5H and $2502H$ holds 92H. 1. $A=C5H$ 2. ADD M calculates $C5H+92H=157H$. 3. The Accumulator only holds 8 bits, so it violently truncates to $57H$. The Carry Flag is SET (CY=1). 4. The JNC instruction fails (because there is a carry). 5. INR C executes, making $C=01H$. 6. The program stores $57H$ at $2503H$ and stores $01H$ at $2504H$. The complete 16-bit result is 0157H.