ITPEC FE Subject B April 2024 Question 6

Filling in missing bitwise logic to count the number of set bits (1s) in an 8-bit value.

The function count1 counts how many bits are 1 in an 8-bit input (popcount). It loops i from 1 to 8 and checks each bit using a bitmask.

The correct expression is:

rbyte & (00000001 << (i - 1))

How it works step by step:

• •00000001 << (i - 1) creates a mask with a single 1 at bit position i - 1
• •i = 1 → mask = 00000001 (bit 0)
• •i = 2 → mask = 00000010 (bit 1)
• •...
• •i = 8 → mask = 10000000 (bit 7)
• •rbyte & mask isolates that one bit using bitwise AND
• •If the result ≠ 00000000, the bit is 1, so the counter r increments

Verification with example: count1(11001011) should return 5.
- i=1: 11001011 & 00000001 = 00000001 ≠ 0 → r=1
- i=2: 11001011 & 00000010 = 00000010 ≠ 0 → r=2
- i=3: 11001011 & 00000100 = 00000000 = 0 → skip
- i=4: 11001011 & 00001000 = 00001000 ≠ 0 → r=3
- i=5: 11001011 & 00010000 = 00000000 = 0 → skip
- i=6: 11001011 & 00100000 = 00000000 = 0 → skip
- i=7: 11001011 & 01000000 = 01000000 ≠ 0 → r=4
- i=8: 11001011 & 10000000 = 10000000 ≠ 0 → r=5 ✓

Why not others:
- (b) `<< i` — shifts 1 to 8 positions, skipping bit 0 and going out of range at i = 8
- (c) `<< (i + 1)` — shifts 2 to 9 positions, missing even more bits
- (d), (e), (f) use `|` (OR) — OR does not isolate a bit; it sets the bit, so the result is almost always non-zero regardless of the original value

Key rule: to test whether a specific bit is set, use AND (&) with a bitmask — never OR (|).

AI-generated — may contain errors