C# bitwise NOT, shifts, and promotions — predict the output byte b1 = 0xAB; byte b2 = 0x99; byte temp; temp = (byte)~b2; Console.Write(temp + ' '); temp = (byte)(b1 2); Console.WriteLine(temp);

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Solve this multiple-choice question and choose the correct option.

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Introduction / Context:This snippet combines bitwise NOT (~), left and right shifts, and type promotion rules for byte in C#. Predicting the output requires careful attention to how shift counts work and how casting back to byte truncates. Given Data / Assumptions: b1 = 0xAB (171), b2 = 0x99 (153). byte is promoted to int for operators; cast back to byte truncates to the low 8 bits. For int shifts, only the low 5 bits of the shift count are used (count mod 32). Concept / Approach:Compute each line independently, remembering promotions and shift-count masking. Step-by-Step Solution: 1) ~b2 → ~0x99 = 0x66 (binary inversion of 10011001 is 01100110) → 102 decimal.2) b1 << b2 → shift count = 153 mod 32 = 25; (int)0xAB << 25 pushes all significant bits out of the low byte → when cast to byte, result is 0.3) b2 >> 2 → 0x99 (153) >> 2 = 0x26 (38) for logical effect after cast to byte (byte promoted to int, right shift on positive int is arithmetic; for 153 it matches logical). Verification / Alternative check:Use a small console program to print each intermediate value in hex and decimal; confirm 0x66, 0x00, 0x26. Why Other Options Are Wrong:They miscompute the complement, ignore shift-count masking, or mis-handle the cast back to byte. Common Pitfalls:Forgetting that shifting by a large number is masked mod 32 for int; expecting left shift to wrap instead of truncating when cast back to byte. Final Answer:102 0 38

C# bitwise NOT, shifts, and promotions — predict the output byte b1 = 0xAB; byte b2 = 0x99; byte temp; temp = (byte)~b2; Console.Write(temp + ' '); temp = (byte)(b1 << b2); Console.Write(temp + ' '); temp = (byte)(b2 >> 2); Console.WriteLine(temp);

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