Embedded C Programming Questions and Answers

Answer: A

Whether the first declared field occupies the low or the high bits, and how fields straddle a unit boundary, are implementation defined, so the same declaration maps differently on two compilers. Masks and shifts on an explicit unsigned type are portable where bit fields are not.

Answer: B

Correct answer: not less

The usual arithmetic conversions turn the signed operand into unsigned when the two have equal rank, so \( -1 \) becomes 4,294,967,295 and is not less than 1. The code compiles and its behaviour is perfectly defined - it simply does something other than what it appears to say, which is why compilers warn about mixed comparisons.

Answer: C

Both operands are promoted to int before the addition, so the sum is 300 with no wrap and the division gives 150, which fits in a byte. Had the arithmetic been done in eight bits the sum would have wrapped to 44 and the answer would have been 22.

Answer: D

Unsigned arithmetic wraps by definition; signed overflow is undefined, and modern optimisers exploit that to assume it cannot happen. Code that relies on signed wrapping can therefore be transformed in surprising ways.

Answer: A

Correct answer: the counter can never reach 300 and the loop never ends

A uint8_t holds 0 to 255, so after 255 it wraps to 0 and the condition remains true for ever. The comparison itself is well defined, since the counter is promoted to int; the fault is that the type cannot represent the bound, which is a common and hard-to-spot error in constrained code.

Answer: B

Signed overflow is undefined, so nothing is guaranteed, although two's complement hardware almost always produces minus 128. Writing code that depends on it is what makes a program fragile under a change of compiler or optimisation level.

Answer: C

A core without a floating point unit emulates every operation in software at tens of cycles each, whereas fixed point arithmetic is ordinary integer work. The costs are a limited dynamic range and the discipline of tracking the scaling.

Answer: D

Correct answer: \( -1 \) to just under \( +1 \)

Q15 places the binary point immediately below the sign bit, so the 16-bit integer is read as a fraction: the range runs from \( -1 \) to \( 1 - 2^{-15} \). The format is used where fractional arithmetic is needed on a processor with no floating point unit.

Answer: A

Correct answer: 16384

The integer stored is the fraction multiplied by \( 2^{15} \): \[ 0.5 \times 32768 = 16384. \] Note that \( +1 \) itself cannot be represented, since \( 1.0 \times 32768 = 32768 \) exceeds the range of a signed 16-bit value - which is why the upper bound is just below one.

Answer: B

The fractional bits add, so the product of two Q15 values has thirty fractional bits and needs a 32-bit accumulator. Shifting right by fifteen returns it to Q15, and doing so before the multiplication instead loses the precision the accumulator preserved.