n no of array in c using malloc

1#include <stdio.h>
2
3int main()
4{
5    printf("\n\n\t\tStudytonight - Best place to learn\n\n\n");
6    int n, i, *ptr, sum = 0;
7
8    printf("\n\nEnter number of elements: ");
9    scanf("%d", &n);
10
11    // dynamic memory allocation using malloc()
12    ptr = (int *) malloc(n*sizeof(int));
13
14    if(ptr == NULL) // if empty array
15    {
16        printf("\n\nError! Memory not allocated\n");
17        return 0;   // end of program
18    }
19
20    printf("\n\nEnter elements of array: \n\n");
21    for(i = 0; i < n; i++)
22    {
23        // storing elements at contiguous memory locations
24        scanf("%d", ptr+i);    
25        sum = sum + *(ptr + i);
26    }
27
28    // printing the array elements using pointer to the location
29    printf("\n\nThe elements of the array are: ");
30    for(i = 0; i < n; i++)
31    {
32        printf("%d  ",ptr[i]);    // ptr[i] is same as *(ptr + i)
33    }
34
35    /* 
36        freeing memory of ptr allocated by malloc 
37        using the free() method
38    */
39    free(ptr);
40
41    printf("\n\n\t\t\tCoding is Fun !\n\n\n");
42    return 0;
43}