What will be the output for the following code?1234567891011#include<stdio.h>#include<stdlib.h>int main() { int *arr = (int*) calloc(3, sizeof(int)); if (arr != NULL) { arr[0] = 10; arr[1] = 20; arr[2] = 30; } printf("%d", arr[1]); free(arr); return 0; }

The output of the code will be 20.

Here's the step by step explanation:

1. The code begins by including the necessary libraries. stdio.h for standard input/output functions and stdlib.h for functions involving memory allocation, control, etc.

2. The main() function is declared, which is the entry point of the program.

3. Inside the main() function, a pointer arr is declared and memory for 3 integers is allocated using the calloc() function. calloc() initializes the allocated memory to zero.

4. A condition is checked to see if the memory allocation was successful. If arr is not NULL, it means the memory allocation was successful.

5. If the memory allocation was successful, the first, second, and third elements of the array pointed to by arr are set to 10, 20, and 30 respectively.

6. The second element of the array (arr[1]) is printed using printf(). Since arrays in C are 0-indexed, arr[1] refers to the second element, which is 20.

7. The allocated memory is freed using free(). This is a good practice to prevent memory leaks.

8. The program returns 0, indicating successful execution.

So, the output of the code will be 20.