Operator precedence with bitwise and logical operators: predict the three integers printed. #include int main() { int i = 4, j = 8; printf("%d, %d, %d ", i | j & j | i, i | j && j | i, i ^ j); return 0; }

Introduction / Context:This example tests detailed precedence and associativity across bitwise AND (&), XOR (^), OR (|), and logical AND (&&). Knowing that bitwise operators have higher precedence than logical && and that & binds more tightly than | is essential.

Given Data / Assumptions:

• i = 4 (binary 0100), j = 8 (binary 1000).
• Precedence (high → low among these): & then ^ then | then &&.
• All bitwise operators associate left-to-right.

Concept / Approach:Parse each expression with precedence rules, reduce bitwise operations to integers, then evaluate logical && which yields 0 or 1. Finally, print the results.

Step-by-Step Solution:Expr1: i | j & j | i → i | (j & j) | i → 4 | 8 | 4 = 12.Expr2: i | j && j | i → (i | j) && (j | i) because | has higher precedence than && → (12) && (12) → 1.Expr3: i ^ j → 0100 ^ 1000 = 1100 → 12.

Verification / Alternative check:Add parentheses explicitly to match the parsing above and re-run; you will see identical outputs: 12, 1, 12.

Why Other Options Are Wrong:They assume different precedence (e.g., evaluating && earlier) or miscompute bitwise results. None match the correct parsing sequence defined by the C standard.

Common Pitfalls:Believing logical && outranks bitwise |, or forgetting that nonzero integers are “true” for logical operators so 12 && 12 is 1, not 12.

Final Answer:12, 1, 12