Evaluate the output of this C program that uses logical negation: #include<stdio.h> int main() { int x = 10, y = 20; if (!(!x) && x) printf("x = %d\n", x); else printf("y = %d\n", y); return 0; }

Introduction / Context:
This item checks how logical negation interacts with nonzero integers in C. Any nonzero value is considered true. Applying ! flips truthiness, not the numeric magnitude (except reducing truthy to 0 and false to 1).

Given Data / Assumptions:

• x = 10 (nonzero → true), y = 20.
• Expression: !(!x) && x.
• Short-circuit semantics of && apply.

Concept / Approach:
!x converts any nonzero to 0 and zero to 1. Thus !x becomes 0, and !(!x) becomes 1. The second operand of && is x (which is nonzero). Therefore, 1 && 10 is true, so the if branch executes, printing the current numeric value of x (10).

Step-by-Step Solution:

Verification / Alternative check:
Replace x by 0 and re-run to see the else branch. Replace x by −5 and note that it is still treated as true.

Why Other Options Are Wrong:

Common Pitfalls:
Confusing logical truthiness with numeric value; assuming !10 creates −10 (it does not).

Final Answer:
x = 10.