Which bitwise operator is suitable for turning off a particular bit in a number ?
& operator
Clearing a bit means ANDing with a mask that holds 0 in that position and 1 everywhere else, so the operator is the bitwise AND — option 2.
x = x & ~(1 << n);
Why AND is the operator that clears. Look at what AND does to a single bit:
| Bit | Mask | Result |
|---|---|---|
| 0 | 1 | 0 — unchanged |
| 1 | 1 | 1 — unchanged |
| 0 | 0 | 0 — cleared |
| 1 | 0 | 0 — cleared |
A mask bit of 1 leaves the data bit alone; a mask bit of 0 forces it to zero. So a mask of all ones with a single zero clears exactly one bit and preserves the rest — which is the whole requirement.
Building the mask. 1 << n gives a word with only bit n set; the complement operator ~ inverts it to all ones except bit n. To clear bit 3 of a byte: 1 << 3 is 00001000, and ~ makes it 11110111, which ANDed with the data leaves everything except bit 3 untouched.
The complete set of bit manipulations follows the same pattern and is worth holding together:
| Operation | Operator | Idiom |
|---|---|---|
| Clear a bit | & | x &= ~(1<<n) |
| Set a bit | | | x |= (1<<n) |
| Toggle a bit | ^ | x ^= (1<<n) |
| Test a bit | & | if (x & (1<<n)) |
Why the other options fail. The distinction between & and && is the real point of the question. && is the logical AND: it treats each operand as a single true-or-false value and yields 1 or 0, so 12 && 10 is 1, whereas 12 & 10 is 8 — it discards the bit pattern entirely. || is the logical OR, and its bitwise counterpart | would set bits rather than clear them. ! is logical negation, which turns any non-zero value into 0; the bitwise complement is ~.
Where this matters in practice : writing to a port register without disturbing the other pins, masking off status flags, and clearing an interrupt bit — all embedded work where a read-modify-write must touch one bit only.
Hence, the operator is the & operator.
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