If the 8085 adds 87H and 79H, specify the contents of the accumulator and the status of the S, Z, and CY flag?
18. If the 8085 adds 87H and 79H, specify the contents of the accumulator and the status of the S, Z, and CY flag?
To add two numbers in the Intel 8085 microprocessor, you would use the ADD instruction. The ADD instruction adds the specified operand to the contents of the accumulator (Acc). Let's calculate the result of adding 87H and 79H using hexadecimal addition:
Here's the breakdown of the addition:
Carry from the lower nibble (4 bits) of 7 + 9 = 10, so there's a carry to the higher nibble.
Carry from the higher nibble of 7 + 8 = F, so there's a carry to the higher byte.
The result is 100H.
Now let's analyze the flags:
- Carry Flag (CY): The carry flag will be set because there was a carry from the most significant bit during the addition. Therefore, CY = 1.
- Zero Flag (Z): The zero flag will be cleared because the result is not zero (it's 100H), so Z = 0.
- Sign Flag (S): The sign flag will be set because the most significant bit of the result (1) is 1 (indicating a negative value in two's complement representation), so S = 1.
The contents of the accumulator will hold the result of the addition, which is 100H.
To summarize:
Accumulator (Acc) = 100H
Carry Flag (CY) = 1 (set)
Zero Flag (Z) = 0 (cleared)
Sign Flag (S) = 1 (set)
Please note that in two's complement representation, the most significant bit (MSB) being 1 indicates a negative value, and in unsigned representation, it indicates a value greater than 127 (decimal). The flags are used to interpret the result and perform conditional branching or decision-making in program flow.
