C short-circuiting with pre-increment side effects: what does this program print? #include<stdio.h> int main() { int i=-3, j=2, k=0, m; m = ++i && ++j || ++k; printf("%d, %d, %d, %d ", i, j, k, m); return 0; }

Introduction / Context:
This problem blends operator precedence (&& before ||), short-circuit evaluation, and side effects of the pre-increment operator. Mastering the order of evaluation and when increments occur is essential for correct reasoning about C expressions.

Given Data / Assumptions:

• Initial values: i = -3, j = 2, k = 0.
• Expression: m = (++i && ++j) || ++k due to precedence.
• Short-circuit rules: stop evaluating && on false left; stop evaluating || on true left.

Concept / Approach:
Evaluate ++i, then ++j only if needed by &&, and ++k only if the left side of || is false. Nonzero values count as true. Carefully track variable updates as you go.

Step-by-Step Solution:

Verification / Alternative check:
Insert prints after each step or separate the expression into temporaries to see the same final state, confirming short-circuit behavior prevents ++k from executing.

Why Other Options Are Wrong:

Common Pitfalls:
Forgetting precedence and imagining left-to-right increments without considering short-circuit; assuming negative values are false (they are true if nonzero).

Final Answer:
-2, 3, 0, 1.