Topic: Microprocessors
How is the status of the carry, auxiliary carry and parity flag affected if write instruction?
MOV A, #9C
ADD A, #64H
- A. $CY=0, AC=0, P=0$
- B. $CY=1, AC=1, P=0$
- C. $CY=0, AC=1, P=0$
- D. $CY=1, AC=1, P=1$
Answer: B. $CY=1, AC=1, P=0$
Explanation: 9CH = 1001 1100
64H = 0110 0100
Addition: 1001 1100 + 0110 0100 = 1 0000 0000 (Result is 00H with a carry).
Since there is a carry out of D7, CY=1.
There is a carry from D3 to D4, so AC=1.
For 8051, P=1 when there is an odd number of 1's in the accumulator. Since 00H has zero 1's, P=0.