Bit operations in C# (Byte): turn OFF the 4th bit from the right (bit value 8) without changing other bits. Which statement performs this correctly?

Introduction / Context:
In C# bit manipulation, clearing (turning OFF) a specific bit in a byte is a common task. This question checks whether you can choose the correct bit-mask and operator to clear the 4th bit from the right (counting 1, 2, 3, 4 → value 8) without disturbing other bits.

Given Data / Assumptions:

• n is of type byte (0–255).
• We must clear the 4th bit (value 8) and keep all other bits unchanged.
• We are using C#'s bitwise operators with hexadecimal masks.

Concept / Approach:
To turn OFF a single bit, AND the value with a mask that has 0 in the target position and 1s elsewhere. For the 4th bit (value 8), the mask is 11110111 in binary, which is 0xF7 in hex. The operation is: n = n & 0xF7

Step-by-Step Solution:

Verification / Alternative check:
Try n = 0xFF (255). After n & 0xF7, result is 0xF7 (247), confirming only that bit was cleared.

Why Other Options Are Wrong:

• A: && is a Boolean operator, not bitwise.
• B: & 16 clears the 5th bit (value 16), not the 4th.
• D: HexF7 is not valid C# syntax.
• E: n = n & 8 zeroes every other bit, not just the 4th.

Common Pitfalls:
Confusing Boolean operators (&&, ||) with bitwise (&, |), and miscounting bit positions (remember that the 4th bit from the right has value 8).

Final Answer:
n = n & 0xF7