Finding even and odd numbers in a set of elements dynamically using C language

Problem

To compute sum of even numbers and odd numbers in a set of elements using dynamic memory allocation functions.

Solution

In this program, we are trying to find even and odd numbers in a set of numbers.

The logic used to find even numbers in a set elements is given below −

for(i=0;i<n;i++){
   if(*(p+i)%2==0) {//checking whether no is even or not
      even=even+*(p+i); //calculating sum of even all even numbers in a list
   }
}

The logic used to find odd numbers in a set elements is given below −

for(i=0;i<n;i++){
   if(*(p+i)%2==0) {//checking number is even or odd
      even=even+*(p+i);
   }
   Else {//if number s odd enter into block
      odd=odd+*(p+i); //calculating sum of all odd numbers in a list
   }
}

Example

#include<stdio.h>
#include<stdlib.h>
void main(){
   //Declaring variables, pointers//
   int i,n;
   int *p;
   int even=0,odd=0;
   //Declaring base address p using malloc//
   p=(int *)malloc(n*sizeof(int));
   //Reading number of elements//
   printf("Enter the number of elements : ");
   scanf("%d",&n);
   /*Printing O/p -
   We have to use if statement because we have to check if memory
   has been successfully allocated/reserved or not*/
   if (p==NULL){
      printf("Memory not available");
      exit(0);
   }
   //Storing elements into location using for loop//
   printf("The elements are : \n");
   for(i=0;i<n;i++){
      scanf("%d",p+i);
   }
   for(i=0;i<n;i++){
      if(*(p+i)%2==0){
         even=even+*(p+i);
      }
      else{
         odd=odd+*(p+i);
      }
   }
   printf("The sum of even numbers is : %d\n",even);
   printf("The sum of odd numbers is : %d\n",odd);
}

Output

Enter the number of elements : 5
The elements are :
34
23
12
11
45
The sum of even numbers is : 46
The sum of odd numbers is : 79