A=0x61, C=0 (arithmetic left shift preserving sign bit)

A=0x87, C=unchanged (RL rotates left without involving carry: 1100 0011 → 1000 0111, bit 7 wraps to bit 0)

A=0x86, C=1 (standard left shift operation with carry out)

A=0x87, C=1 (bit 7 shifted into carry flag)

A=1 (infinity approximated as maximum finite value), B=0, OV=1

A=0 (quotient when dividing by zero is defined as zero), B=18 (remainder equals dividend)

A=undefined (garbage), B=undefined (garbage), OV=1 (division by zero error detected)

A=18 (original dividend), B=0 (original divisor), unchanged because operation cannot proceed

A=0xFF, C=1 (saturated at maximum 8-bit value with overflow indication)

A=0x00, C=1 (80H + 80H = 100H, low byte is 00H, bit 8 goes to carry)

A=0x00, C=0 (addition wraps to zero with no carry)

A=0x80, C=0 (addition canceled out and returned to original)

0x55, because XRL performs bitwise XOR where each bit is inverted (AAH = 1010 1010, FFH = 1111 1111, XOR = 0101 0101 = 55H)

0xAA, unchanged because XOR with all 1s preserves the original value

0xFF, by setting all bits to 1 through the XOR operation

0x00, by clearing all bits through exclusive OR

The result will be 0x04 instead of the expected 0x05, because SUBB subtracts both the operand AND the carry flag (15H - 10H - 1 = 04H)

The operation will fail and set an error flag because C must be 0 for SUBB

The result will be correct (0x05) because SUBB automatically clears C before operation

C=1 indicates the previous result was negative, so the current subtraction adds instead

The addition operation failed and the result is invalid

The 16-bit result fits within 16 bits (no 17-bit overflow); if C=1, the true result would be 17 bits (0x1xxxx)

The low byte addition produced no carry, so the high byte operation was unnecessary

The subtraction borrow from the previous operation is still set

0xC5, unchanged because SWAP requires two operands

0xE7, by performing arithmetic on the nibbles

0x3A, by inverting bits during the swap operation

0x5C, because SWAP exchanges the upper nibble (C) with the lower nibble (5), resulting in 0101 1100

Use repeated addition (add the number to itself 15 times), which avoids multiplication entirely

Use MUL AB instruction with B equals 16, which takes 48 clocks but is more reliable than shifts

Use software multiplication algorithm from Practical 26, which is more accurate than hardware

Perform four consecutive left shifts using RL A (4 machine cycles total), since each left shift multiplies by 2 and 2 to the power of 4 equals 16

0x0A, because ANL performs bitwise AND, preserving only the lower nibble (9AH = 1001 1010, 0FH = 0000 1111, AND = 0000 1010 = 0AH)

0x9A, unchanged because AND doesn't modify the accumulator

0x90, by clearing the lower nibble to all 0s

0x9F, by setting the lower nibble bits to all 1s

Because ADD handles unsigned numbers while ADDC handles signed numbers

Because ADDC requires the carry flag to be cleared first, which ADD does automatically

Because ADD clears and ignores the carry flag, while ADDC includes the carry from the previous byte addition, enabling the carry chain necessary for multi-precision arithmetic

Because ADD is faster than ADDC, so we use it for the first byte to optimize performance