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

 Live Demo

#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 : 
");    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
",even);    printf("The sum of odd numbers is : %d
",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