Evaluate bitwise precedence and grouping. For i = 4 and j = 8, what is printed by the following expression sequence using & (AND), ^ (XOR), and | (OR)? #include<stdio.h> int main() { int i = 4, j = 8; printf("%d, %d, %d\n", i | j & j | i, i | j & j | i, i ^ j); return 0; }

Introduction / Context:
The task tests operator precedence for bitwise operators in C: & has higher precedence than ^, which has higher precedence than |. Understanding this avoids accidental mis-grouping.

Given Data / Assumptions:

• i = 4 (binary 0100), j = 8 (binary 1000).
• Expression: i | j & j | i, evaluated twice, then i ^ j.

Concept / Approach:
Evaluate j & j first, then apply |. For XOR, evaluate directly. No parentheses are present, so rely on precedence rules (& before ^ before |) and left-to-right association within the same precedence level for | and ^.

Step-by-Step Solution:

Verification / Alternative check:
Compute using binary: 0100 | 1000 = 1100 (12); 0100 ^ 1000 = 1100 (12). Both confirm the same decimal result.

Why Other Options Are Wrong:

• 112, 1, 12 and 32, 1, 12 and -64, 1, 12: These ignore precedence and invent unintended groupings or arithmetic.

Common Pitfalls:
Assuming all bitwise operators have the same precedence or that evaluation is strictly left-to-right without considering operator ranks. 

Final Answer:
12, 12, 12