Bit Manipulation - Power of Four
Given the output of the bitmask check (n & 0x55555555) == 0 is false, and (n & (n - 1)) == 0 is true, what can you infer about the original input n?
Given the output of the bitmask check (n & 0x55555555) == 0 is false, and (n & (n - 1)) == 0 is true, what can you infer about the original input n?
hard🔄 Reverse Engineer Q9 of Q15
Step-by-Step Solution
Solution:
Step 1: Interpret (n & (n - 1)) == 0 is true
This means n has exactly one set bit (power of two).Step 2: Interpret (n & 0x55555555) == 0 is false
Means the set bit is NOT in an even position, so not a power of four.Final Answer:
Option B -> Option BQuick Check:
Single set bit but wrong bit position -> power of two but not four [OK]
Quick Trick: Single set bit but mask fails -> power of two not four [OK]
Common Mistakes:
MISTAKES- Confusing mask meaning
- Assuming true means power of four
Trap Explanation:
PITFALL- Candidates often miss bit position check meaning in mask result.
Interviewer Note:
CONTEXT- Tests reverse reasoning from bitmask outputs
Master "Power of Four" in Bit Manipulation
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