how to use malloc to get a array code example

Example 1: how to dynamically allocate array size in c

// declare a pointer variable to point to allocated heap space
int    *p_array;
double *d_array;

// call malloc to allocate that appropriate number of bytes for the array

p_array = (int *)malloc(sizeof(int)*50);      // allocate 50 ints
d_array = (int *)malloc(sizeof(double)*100);  // allocate 100 doubles


// use [] notation to access array buckets 
// (THIS IS THE PREFERED WAY TO DO IT)
for(i=0; i < 50; i++) {
  p_array[i] = 0;
}

// you can use pointer arithmetic (but in general don't)
double *dptr = d_array;    // the value of d_array is equivalent to &(d_array[0])
for(i=0; i < 50; i++) {
  *dptr = 0;
  dptr++;
}

Example 2: n no of array in c using malloc

#include <stdio.h>

int main()
{
    printf("\n\n\t\tStudytonight - Best place to learn\n\n\n");
    int n, i, *ptr, sum = 0;

    printf("\n\nEnter number of elements: ");
    scanf("%d", &n);

    // dynamic memory allocation using malloc()
    ptr = (int *) malloc(n*sizeof(int));

    if(ptr == NULL) // if empty array
    {
        printf("\n\nError! Memory not allocated\n");
        return 0;   // end of program
    }

    printf("\n\nEnter elements of array: \n\n");
    for(i = 0; i < n; i++)
    {
        // storing elements at contiguous memory locations
        scanf("%d", ptr+i);    
        sum = sum + *(ptr + i);
    }

    // printing the array elements using pointer to the location
    printf("\n\nThe elements of the array are: ");
    for(i = 0; i < n; i++)
    {
        printf("%d  ",ptr[i]);    // ptr[i] is same as *(ptr + i)
    }

    /* 
        freeing memory of ptr allocated by malloc 
        using the free() method
    */
    free(ptr);

    printf("\n\n\t\t\tCoding is Fun !\n\n\n");
    return 0;
}