In C, evaluate logical expressions and operator precedence. What is the output of the following program? #include<stdio.h> int main() { int i=4, j=-1, k=0, w, x, y, z; w = i || j || k; x = i && j && k; y = i || j && k; z = i && j || k; printf("%d, %d, %d, %d ", w, x, y, z); return 0; }

Introduction / Context:
C's logical operators use short-circuit evaluation and defined precedence: && (logical AND) has higher precedence than || (logical OR). Any nonzero integer is treated as true, and zero is false. This problem checks your command of precedence and short-circuit semantics.

Given Data / Assumptions:

• i = 4 (true), j = -1 (true), k = 0 (false).
• Operator precedence: && before ||.
• Logical result values are 0 or 1.

Concept / Approach:
Evaluate expressions respecting precedence, then apply short-circuiting: for A || B, if A is true, B is not evaluated; for A && B, if A is false, B is not evaluated. Here, all operands are simple variables, so no side effects complicate evaluation.

Step-by-Step Solution:

Verification / Alternative check:
Manually substituting 1 for nonzero and 0 for zero reproduces the results, confirming 1, 0, 1, 1.

Why Other Options Are Wrong:

Common Pitfalls:
Forgetting that && binds tighter than ||; assuming nonzero negative values are false (they are true in C).

Final Answer:
1, 0, 1, 1.