TEZPUR 2025 — Computer PYQ

The correct answer is (b) 1.

To understand this result, we must analyze the behavior of the if condition using short-circuit evaluation and the decrement operator (--).

1. Logical AND (&&) Short-Circuiting: The && operator evaluates from left to right. If the left operand is false (0), the entire expression becomes false, and the right operand is not evaluated.

2. Step-by-Step Execution:

   • Initialize x = 1.

   • Evaluate the if condition: (x && --x).

   • The first part, x, is 1, which is true in C.

   • Because the left side is true, the compiler must evaluate the right side, --x.

   • The decrement operator --x (pre-decrement) reduces x by 1 immediately. So, x becomes 0.

   • The condition now effectively evaluates (1 && 0), which is 0 (false).

3. Final Outcome:

   • Since the if condition resulted in 0 (false), the code block x += 2; is skipped.

   • The value of x remains 0 after the decrement, but wait—let's re-verify the post-evaluation.

   • Actually, x was 1, then --x turned it to 0. The condition is false. The code exits the if block. x is currently 0. The printf will print 0.

Correction: Looking at the code: x starts at 1. if(1 && 0) is 0. The if body is not executed. x is 0. The answer is (a) 0.

We can represent the logical operation as:

$$\text{Condition}=x\wedge (\text{–}x)$$

Given $x=1$:

$$1\wedge 0=0$$

Since the condition is false, the assignment x += 2 never occurs. Thus, the final value of $x$ is:

$$x_{final}=0$$