C logical OR (||) evaluation: what does this program output? #include<stdio.h> int main() { int x=12, y=7, z; z = x!=4 || y == 2; printf("z=%d ", z); return 0; }

Introduction / Context:Logical operators in C convert relational results to either 0 (false) or 1 (true). This simple example reinforces that comparisons yield 0 or 1 and that the logical OR operator short-circuits once a true operand is found.

Given Data / Assumptions:

• x = 12, y = 7.
• Expression: z = (x != 4) || (y == 2).
• Standard C semantics for relational and logical operators.

Concept / Approach:Evaluate the left comparison first: x != 4 is true (12 is not 4), which yields 1. Since || short-circuits, the right side does not affect the final result; z is set to 1. The printf prints z as “1”.

Step-by-Step Solution:

Verification / Alternative check:Replace y with any value; as long as x != 4 remains true, z stays 1. Conversely, set x to 4 and y to 2 to see z remain 1 due to the right operand.

Why Other Options Are Wrong:

Common Pitfalls:Thinking relational operators return operands themselves; in C99+ the result is guaranteed 0 or 1 for these logical forms.

Final Answer:z=1.