Question:
What is the state of carry, auxiliary carry and parity flags after execution of following instructions:
MOV A, #3F
ADD A, #BE
What is the state of carry, auxiliary carry and parity flags after execution of following instructions:
MOV A, #3F
ADD A, #BE
ADD A, #BE
Show Hint
Parity = 1 for even number of 1s, else 0.
Updated On: Jul 2, 2026
- CY=0, AC=0, P=0
- CY=1, AC=1, P=0
- CY=0, AC=1, P=0
- CY=1, AC=1, P=1
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
Flags in 8051 are affected by arithmetic operations.
Step 1: Convert values to binary \[ 3F = 0011\ 1111 \] \[ BE = 1011\ 1110 \]
Step 2: Perform addition \[ 00111111 + 10111110 = 11111101 \]
Step 3: Carry flag (CY) Since sum exceeds 8 bits → Carry = 1
Step 4: Auxiliary Carry (AC) Carry from lower nibble (4 bits) → AC = 1
Step 5: Parity flag (P) Result = 11111101 → number of 1s = 7 (odd) So parity = 0
Conclusion: \[ \boxed{CY=1,\ AC=1,\ P=0} \]
Final Answer: \[ \boxed{(B)} \]
Step 1: Convert values to binary \[ 3F = 0011\ 1111 \] \[ BE = 1011\ 1110 \]
Step 2: Perform addition \[ 00111111 + 10111110 = 11111101 \]
Step 3: Carry flag (CY) Since sum exceeds 8 bits → Carry = 1
Step 4: Auxiliary Carry (AC) Carry from lower nibble (4 bits) → AC = 1
Step 5: Parity flag (P) Result = 11111101 → number of 1s = 7 (odd) So parity = 0
Conclusion: \[ \boxed{CY=1,\ AC=1,\ P=0} \]
Final Answer: \[ \boxed{(B)} \]
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