How does the program find the quotient and the remainder?

It subtracts the divisor while the dividend is not less than it; the count is the quotient, the leftover the remainder

It shifts the dividend right once for each bit in the divisor

It uses the 8085 DIV instruction to get both results

It adds the divisor to itself until the sum equals the dividend

In the loop, what do CMP B and JC DONE do?

CMP B sets CY = 1 if A < B, so JC DONE stops when no more subtraction is possible

CMP B subtracts B from A, and JC loops while there is a carry

CMP B sets Z = 1 if A < B, and JC jumps on zero

CMP B sets CY = 1 if A > B, so the loop stops when the dividend is larger

Why does the program run INR B followed by DCR B?

To set Z if B is 0 (B is unchanged), so a zero divisor goes to ERROR

To round the divisor up by one before dividing

To clear the carry flag before the first CMP B

To start the quotient at one instead of zero

If 9070H = 64H (100) and 9071H = 07H, what does the program store?

9072H = 02H (remainder) and 9073H = 0EH (quotient)

2070 Magh · 8 marks

8085 ALP: divide two bytes

Short answer

Divide by repeated subtraction of the divisor from the dividend; each successful subtraction increments the quotient. The remaining dividend is the remainder. If divisor is zero, store FFH in both result locations. For 100÷7, quotient = 0EH, remainder = 02H.

Write an assembly language program in 8085 to divide a byte stored in memory location 9070 H by byte stored in 9071 H and store the remainder and quotient at 9072 H and 9073 H respectively.

Method: repeated subtraction. The divisor is subtracted from the dividend as long as the dividend is not smaller than the divisor; each successful subtraction increments the quotient. What is left is the remainder. A divide-by-zero check is included to prevent an endless loop.

        LXI H,9070H
        MOV A,M         ; A = dividend (9070H)
        INX H
        MOV B,M         ; B = divisor (9071H)
        INR B
        DCR B           ; test divisor for zero
        JZ ERROR
        MVI C,00H       ; C = quotient
LOOP:   CMP B           ; CY = 1 if A < B
        JC DONE
        SUB B           ; subtract divisor
        INR C           ; count it
        JMP LOOP
DONE:   INX H           ; HL = 9072H
        MOV M,A         ; remainder
        INX H
        MOV M,C         ; quotient at 9073H
        HLT
ERROR:  INX H           ; divide by zero: store FFH in both
        MVI M,0FFH
        INX H
        MVI M,0FFH
        HLT

Example: 9070H = 64H (100), 9071H = 07H gives quotient 0EH (14) at 9073H and remainder 02H at 9072H.

✓ Answer checked by Claude Opus 5.5 (high effort)

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