Consider the following C program. #include <stdio.h> int main() { int i, j,…
2021
Consider the following C program.
#include <stdio.h>
int main()
{
int i, j, count;
count = 0;
i = 0;
for (j = -3; j <= 3; j++)
{
if ((j >= 0) && (i++))
{
count = count + j;
}
}
count = count + i;
printf("%d", count);
return 0;
}
Answer: B. The program will compile successfully and output 10 when executed. — Key insight: the logical AND expression (j >= 0) && (i++) uses short-circuit evaluation, so i++ is executed only when j >= 0. The i++ expression returns the…
- A.
The program will not compile successfully.
- B.
The program will compile successfully and output 10 when executed.
- C.
The program will compile successfully and output 8 when executed.
- D.
The program will compile successfully and output 13 when executed.
Attempted by 289 students.
Show answer & explanation
Correct answer: B
Key insight: the logical AND expression (j >= 0) && (i++) uses short-circuit evaluation, so i++ is executed only when j >= 0. The i++ expression returns the previous value of i (then increments i), which affects whether the if-body runs.
Initial values: i = 0, count = 0.
For j = -3, -2, -1: j >= 0 is false, so i++ is not evaluated; i stays 0 and count remains 0.
For j = 0: j >= 0 is true, so i++ is evaluated. i++ returns 0 (making the if condition false) but increments i to 1. count stays 0.
For j = 1: i++ returns 1 and i becomes 2. The if condition is true, so count += 1 -> count = 1.
For j = 2: i++ returns 2 and i becomes 3. The if condition is true, so count += 2 -> count = 3.
For j = 3: i++ returns 3 and i becomes 4. The if condition is true, so count += 3 -> count = 6.
After the loop: i = 4 and count = 6. The statement count = count + i makes count = 6 + 4 = 10.
Answer: The program prints 10.
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