C#.NET — Determine the printed output by evaluating bitwise precedence and Convert.ToBoolean. int i = 2, j = i; if (Convert.ToBoolean((i | (j & 5)) & (j - 25 * 1))) Console.WriteLine(1); else Console.WriteLine(0);

Difficulty: Medium

Correct Answer: 0

Explanation:


Introduction / Context:
This question checks precedence of bitwise operators, arithmetic evaluation, and the effect of Convert.ToBoolean on integer values in C#.


Given Data / Assumptions:

  • i = 2, j = 2
  • Expression: Convert.ToBoolean((i | j & 5) & (j - 25 * 1))
  • Console prints 1 if the boolean is true, otherwise 0.


Concept / Approach:
Operator precedence: & (bitwise AND) binds tighter than | (bitwise OR). Arithmetic executes before bitwise. Convert.ToBoolean(x) is true for any nonzero integer and false for zero.


Step-by-Step Solution:

Compute j & 5 first: 2 & 5 = 0. Compute i | (that): 2 | 0 = 2. Compute (j - 25 * 1): 2 - 25 = -23. Now evaluate 2 & (-23). In two's complement, -23 has a 0 bit in the 2's place, so 2 & (-23) = 0. Convert.ToBoolean(0) == false; therefore, the else branch runs and prints 0.


Verification / Alternative check:
You can reason by bit positions: value 2 has only the 2's bit set; -23 ends with ...1001 (for -23 it is 1111...1110 1001), making the 2's bit 0; their AND is 0.


Why Other Options Are Wrong:

  • 1: would require a nonzero result after the final &.
  • Compile Error / Run time Error: the code is valid and runs without exceptions.


Common Pitfalls:
Mixing up logical && with bitwise &, and assuming negative & positive is always nonzero. It depends on overlapping set bits.


Final Answer:
0

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