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What will be the output of the program?

#include <stdio.h>

int main()
{
    unsigned int res;
    res = (64 >>(2+1-2)) & (~(1<<2));
    printf("%d\n", res);
    return 0;
}

Answer: A

Correct answer: 32

Take the two halves separately. The shift count evaluates first: \( 2+1-2 = 1 \), so the left operand is \( 64 \gg 1 = 32 \).

The right operand is ~(1<<2), a mask with every bit set except bit 2 - the clear-bit-2 mask.

Now \( 32 = 100000_2 \) has its only bit in position 5, which the mask leaves alone, so the AND changes nothing:

\[ 32 \mathbin{\&} \mathtt{\sim}4 = 32 \]

The masking is a decoy: it would have mattered only if the value had bit 2 set.

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