A bit is to be cleared without disturbing the others. The correct line is
Answer: C
Correct answer: reg &= ~(1u << n);
Shifting one into position and inverting gives a mask with a zero at that bit and ones everywhere else; ANDing with it clears the bit and leaves the rest untouched. OR would set the bit, XOR would toggle it, and plain assignment would destroy every other bit in the register.
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What does this expression evaluate to when bit 3 of reg is set?
(reg >> 3) & 1u
Answer: D
Shifting brings the wanted bit to position zero and the mask discards everything above it, so the result is one or zero. Testing with reg & 8 instead yields eight or zero, which is also true or false in C but is not the same value.
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The value of y after this runs is
uint8_t y = 1; y <<= 7;
Answer: A
Shifting one left seven places puts it in the top bit of a byte, which is 0x80. One more shift would push it out and leave zero, and a shift by eight or more of an 8-bit value is undefined behaviour in principle and unhelpful in practice.
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Shifting a 32-bit unsigned value left by 32 places in C is
Answer: B
A shift count at or beyond the width of the promoted operand is undefined, and real processors do different things: some give zero and some use only the low five bits of the count and shift by nothing at all. Both are conforming, which is why the case must be avoided.
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A mask to keep the low six bits of a byte is
Answer: C
Correct answer: 0x3F
Six bits set is 0011 1111, which is \( 0x3F \). A quick check: a mask of n low bits is \( 2^n - 1 \), so \( 64 - 1 = 63 \), and 63 in hexadecimal is \( 0x3F \). The distractor \( 0x60 \) confuses the count of bits with their positions.
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On a 32-bit target with natural alignment, what is sizeof(struct s)?
struct s {
uint8_t a;
uint32_t b;
uint8_t c;
};
Answer: D
Three bytes of padding follow a so that b lands on a four-byte boundary, and three more follow c so the whole structure is a multiple of its strictest alignment. Reordering the members as b, a, c would bring it down to eight.
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Reordering that structure's members largest first would reduce its size to
Answer: A
With b first, a and c then occupy two of the following bytes and two bytes of tail padding complete the four-byte multiple, giving eight. Declaring members in descending order of size is a cheap habit that saves memory in any structure instantiated many times.
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Applying a packed attribute to a structure removes the padding but
Answer: B
On a core that cannot fetch an unaligned word the compiler must generate several byte accesses instead of one, and on some cores an unaligned access faults outright. Packing is for matching an external format, not for saving memory in internal structures.
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Casting a byte buffer to a structure pointer to parse a network packet is unsafe chiefly because of
Answer: C
Correct answer: alignment and padding assumptions
The compiler inserts padding to align members, and the layout it chooses need not match the packing of the network protocol. Worse, a byte buffer may not be aligned as the structure requires, and an unaligned access faults on some processors. Parsing field by field, or a packed structure with the endianness handled explicitly, is the safe route.
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A union is used in embedded code chiefly to
Answer: D
Correct answer: view the same storage as more than one type
A union overlays its members on the same bytes, so a 32-bit word can be examined as four bytes or a float inspected bitwise. That reinterpretation is its purpose. It does not reduce alignment - a union takes the strictest requirement of its members - and only one member holds a valid value at a time.
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