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GO Classes DPP | GATE CS | C Programming | Nested Loops

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13 13 votes

What is the output of the following code?

#include <stdio.h>

int main() {
    int i, j, sum = 0;

    for (i = 1; i <= 4; i++) {
        for (j = 1; j <= 4; j++) {
            if (j == i)
                continue;

            if (i + j > 5)
                break;

            sum = sum + i + j;
        }
    }

    printf("%d", sum);
    return 0;
}

1 Answer

4 4 votes

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

For $\texttt{i = 1}$,
$\texttt{j = 1}$ is skipped because $\texttt{j == i}$.
Then values added are $\texttt{1 + 2}$, $\texttt{1 + 3}$, and $\texttt{1 + 4}$.
So, $\texttt{sum = 0 + 3 + 4 + 5 = 12}$.

For $\texttt{i = 2}$,
$\texttt{j = 1}$ adds $\texttt{2 + 1 = 3}$,
$\texttt{j = 2}$ is skipped, and $\texttt{j = 3}$ adds $\texttt{2 + 3 = 5}$.
For $\texttt{j = 4}$, $\texttt{i + j = 6}$, so $\texttt{break}$ occurs.
Now, $\texttt{sum = 12 + 3 + 5 = 20}$.

For $\texttt{i = 3}$,
$\texttt{j = 1}$ adds $\texttt{3 + 1 = 4}$, and $\texttt{j = 2}$ adds $\texttt{3 + 2 = 5}$.
Then $\texttt{j = 3}$ is skipped because $\texttt{j == i}$.
For $\texttt{j = 4}$, $\texttt{i + j = 7}$, so $\texttt{break}$ occurs.
Now, $\texttt{sum = 20 + 4 + 5 = 29}$.

For $\texttt{i = 4}$,
$\texttt{j = 1}$ adds $\texttt{4 + 1 = 5}$.
For $\texttt{j = 2}$, $\texttt{i + j = 6}$, so $\texttt{break}$ occurs.
Now, $\texttt{sum = 29 + 5 = 34}$.

Therefore, the output is $\texttt{34}$.

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