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What does this code print?* print_something - function to print something** Return: Always 0**/int print_something(int num){ int i, j; for (i = 0; i < num + 1; i++) { for (j = 0; j < num + 1; j++) { printf("%d", i * j); if (j < num) { printf(", "); } } printf("\n"); } return (0);}The numbers 0 to n, n timesThe n times table, excluding zeroThe n times table, starting with 0I don't know

Question

What does this code print?* print_something - function to print something** Return: Always 0**/int print_something(int num){ int i, j; for (i = 0; i < num + 1; i++) { for (j = 0; j < num + 1; j++) { printf("%d", i * j); if (j < num) { printf(", "); } } printf("\n"); } return (0);}The numbers 0 to n, n timesThe n times table, excluding zeroThe n times table, starting with 0I don't know

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Solution

The code prints the n times table, starting with 0.

Here's a step-by-step explanation:

  1. The function print_something(int num) is defined, which takes an integer num as an argument.

  2. Two integer variables i and j are declared.

  3. A for loop is initiated with i ranging from 0 to num (inclusive).

  4. Inside this loop, another for loop is initiated with j ranging from 0 to num (inclusive).

  5. Inside the inner loop, the product of i and j is printed. If j is less than num, a comma and a space are printed.

  6. After the inner loop finishes, a newline character is printed.

  7. This process repeats until the outer loop finishes.

  8. The function then returns 0.

So, for a given num, this function prints a multiplication table from 0 to num, with each row and column representing the numbers from 0 to num.

This problem has been solved

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