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Given the intermediate state sum_ = 0xFFFFFFFD and carry = 4 in the recursive bitwise sum, what were the original inputs a and b?

hard🔄 Reverse Engineer Q9 of Q15
Bit Manipulation - Sum of Two Integers Without + Operator
Given the intermediate state sum_ = 0xFFFFFFFD and carry = 4 in the recursive bitwise sum, what were the original inputs a and b?
Aa = 0xFFFFFFFE, b = 3
Ba = 0xFFFFFFFD, b = 4
Ca = 1, b = 2
Da = 0xFFFFFFFC, b = 5
Step-by-Step Solution
Solution:
  1. Step 1: Recall sum_ = (a XOR b) & MASK and carry = ((a AND b) << 1) & MASK

    Given sum_ and carry, original a and b satisfy these equations.
  2. Step 2: Recognize sum_ and carry are intermediate results, so a and b are sum_ and carry from previous call

    Thus, a = 0xFFFFFFFD, b = 4.
  3. Final Answer:

    Option B -> Option B
  4. Quick Check:

    Intermediate sum_ and carry correspond to inputs for next recursion [OK]
Quick Trick: Intermediate sum_ and carry become next inputs [OK]
Common Mistakes:
MISTAKES
  • Confusing intermediate results with original inputs
Trap Explanation:
PITFALL
  • Candidates mistake intermediate XOR and carry as final inputs.
Interviewer Note:
CONTEXT
  • Tests deep understanding of recursive state transitions
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