Bitwise Operators Questions and Answers

Answer: A

Correct answer: True

AND with a mask isolates exactly the bits the mask selects, and a mask may name as many positions as wanted. To test whether any of several bits is set, OR the masks together and AND once: if (n & (BIT2 | BIT5)) is true if either is on.

Testing that all of them are on needs a comparison rather than a mere test: if ((n & mask) == mask). That distinction - non-zero means at least one, equality means all - is the practical point behind the question.

Answer: B

Correct answer: False

The bitwise operators require operands of integer type; applying one to a float or double is a compile-time error. To manipulate the bits of a floating-point value the program must first reinterpret it through an integer type of the same width.

Answer: B

Correct answer: Hexadecimal system

Sixteen is a power of two, so each hexadecimal digit corresponds to exactly four binary digits and the conversion is a matter of grouping rather than of arithmetic.

Splitting the number from the right gives 1011 0111 1100 0101, which reads directly as B7C5. Octal shares the property with groups of three bits, but sixteen bits divide evenly into four hexadecimal digits and not into whole octal ones, which is why hex is the convention for byte-oriented data. Decimal is a power of ten and requires repeated division.

Answer: B

Correct answer: & operator

Turning a bit off means forcing it to 0 while leaving the others untouched, and only AND can do that: \( b \mathbin{\&} 0 = 0 \) for any \( b \), while \( b \mathbin{\&} 1 = b \).

The idiom is therefore n &= ~(1 << k) - build a mask with a single 1 in position \( k \), invert it so that position alone is 0, and AND. && and || are logical operators yielding only 0 or 1 and destroying the other bits entirely; ! is logical negation of the whole value.

Answer: A

Correct answer: True

The least significant bit carries the value 1, so it is set precisely for odd numbers. Masking with 1 is therefore a faster parity test than taking the remainder, and unlike the remainder operator it behaves the same for negative values in two’s complement.

Answer: C

Correct answer: 18

The key is operator precedence: the shift operators bind more tightly than the bitwise OR, so the expression groups as (i << 2) | (i >> 1), not as some left-to-right reading.

With \( i = 4 \), which is 00000100:

i << 2 moves the bits two places left, giving 00010000 = 16 - a left shift by \( n \) multiplies by \( 2^{n} \), and \( 4 \times 4 = 16 \).
i >> 1 moves them one place right, giving 00000010 = 2, that is \( 4 / 2 \).

ORing the two combines their set bits, and since they occupy different positions nothing overlaps: 00010000 | 00000010 = 00010010 = \( 16 + 2 = \) 18.

Both shifts read the original value of i; the assignment happens only once the whole right-hand side has been evaluated, so the left shift does not feed the right one.

Answer: A

Correct answer: True

Right-shifting a signed value with the sign bit set is implementation-defined in C: the standard permits either an arithmetic shift, propagating the sign bit so the value stays negative, or a logical shift, bringing in zeros so it becomes a large positive number.

In practice every common compiler chooses the arithmetic shift, but portable code cannot rely on it because the standard does not require it. The safe course is to cast to an unsigned type when the bit pattern is wanted, and to divide when the arithmetic result is wanted.

Answer: A

Correct answer: True

Applying a = a ^ b, then b = a ^ b, then a = a ^ b leaves the two values exchanged, because XOR is its own inverse. The trick fails when both names refer to the same object, since the variable then XORs with itself and becomes zero.

Answer: A

Correct answer: ffff

The statement ~a; computes a value and throws it away. ~ has no side effect on its operand - only an assignment such as a = ~a or the compound form would change a.

So a is still 0xffff when it is printed. Most compilers warn about a statement with no effect, which is the practical defence against this mistake. The intended complement would have been 0x0000, which is the distractor the question offers.

Answer: B

Correct answer: True

Subtracting 1 turns the lowest set bit to 0 and every bit below it to 1, so ANDing with the original leaves that bit and all below it clear. Repeating the operation until n becomes 0 counts the set bits, which is Kernighan’s algorithm.