What will be the output of the following C code? #include int main() int x = 0, y = 2; int z = ~x & y; printf("%d ", z);

Answer: Option B

Understanding Bitwise Operators:
This question tests your understanding of bitwise operators in C. Let's break down the code step by step.

1. `~x`: The tilde (~) is the bitwise NOT operator. It flips all the bits of the variable `x`. Since `x` is 0 (which is represented as all 0s in binary), `~x` will become -1 (represented as all 1s in two's complement).

2. `&y`: The ampersand (&) is the bitwise AND operator. It compares the corresponding bits of two operands. If both bits are 1, the resulting bit is 1; otherwise, it's 0.

3. `~x & y`: We're performing a bitwise AND between `~x` (-1, all 1s) and `y` (2). Let's assume your system uses 32-bit integers (most do). Then 2 is represented as `00000000000000000000000000000010` in binary. The bitwise AND operation will be:
11111111111111111111111111111111 (&) 00000000000000000000000000000010 = 00000000000000000000000000000010
This results in 2.

4. `printf("%d\n", z);` This line prints the value of `z`, which is 2, to the console.

Therefore, the correct answer is B: 2
Important Note: The exact binary representation and the behavior of bitwise NOT might depend slightly on your specific compiler and system architecture, but the principle remains the same.