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GO Classes DPP | GATE CS | C Programming | Loop Control

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5 5 votes

What is the output of the following code?

#include <stdio.h>

int main() {
    int a[5] = {2, 4, 1, 3, 5};
    int i, sum = 0;

    for (i = 0; i < 5; i++) {
        if (a[i] % 2 == 0)
            sum = sum + a[i];
        else {
            sum = sum + i;

            if (sum > 8)
                break;
        }
    }

    printf("%d %d", i, sum);
    return 0;
}
  1. $\texttt{3\ 11}$
     
  2. $\texttt{4\ 16}$
     
  3. $\texttt{5\ 16}$
     
  4. $\texttt{2\ 8}$

2 Answers

2 2 votes

Initially, $\texttt{sum = 0}$.

For $\texttt{i = 0}$, $\texttt{a[0] = 2}$, which is even. So, $\texttt{sum = 0 + 2 = 2}$.

For $\texttt{i = 1}$, $\texttt{a[1] = 4}$, which is even. So, $\texttt{sum = 2 + 4 = 6}$.

For $\texttt{i = 2}$, $\texttt{a[2] = 1}$, which is odd. So, $\texttt{sum = 6 + 2 = 8}$. Since $\texttt{sum > 8}$ is false, the loop continues.

For $\texttt{i = 3}$, $\texttt{a[3] = 3}$, which is odd. So, $\texttt{sum = 8 + 3 = 11}$.

Now $\texttt{sum > 8}$ is true, so $\texttt{break}$ is executed.

Therefore, the loop stops at $\texttt{i = 3}$ and $\texttt{sum = 11}$.

Answer: A. $\texttt{3\ 11}$

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