GATE CS 2021 Set 1 Q37 — Programming in C

The program uses the logical AND operator (&&), which exhibits short-circuit evaluation. In the expression (j >= 0) && (i++), the right-hand side (i++) is evaluated only if the left-hand side (j >= 0) is true.

Let's trace the loop for $j$ from $-3$ to $3$ (initial values: i = 0, count = 0):

• $j = -3, -2, -1$: Since j >= 0 is false, the right operand i++ is not evaluated. i remains $0$, and count remains $0$.
• $j = 0$: j >= 0 is true. i++ is evaluated. The post-increment returns the current value of i ($0$, which is false) and then increments i to $1$. Since the condition is false, count remains $0$.
• $j = 1$: j >= 0 is true. i++ is evaluated. It returns $1$ (true) and increments i to $2$. The condition is true, so count = 0 + 1 = 1.
• $j = 2$: j >= 0 is true. i++ is evaluated. It returns $2$ (true) and increments i to $3$. The condition is true, so count = 1 + 2 = 3.
• $j = 3$: j >= 0 is true. i++ is evaluated. It returns $3$ (true) and increments i to $4$. The condition is true, so count = 3 + 3 = 6.

After exiting the loop, the final values are i = 4 and count = 6.
The statement count = count + i executes:
$$\text{count} = 6 + 4 = 10$$

Thus, the program compiles successfully and outputs 10.