Skip to main content

Program to find sum of even and odd numbers (using one for loop) in C

Instead of using two for loops as in the previous program, only one for loop can be used. Let us see how this can be achieved.

sum= 0;
/* To add all numbers */
for(i=1; i<=n; i++)
sum = sum+i;

It is easy to say that we are computing the sum of all the numbers from 1 to n. To add only even numbers or to add numbers, the code can be changed as shown below:

esum=0;                          /* To store sum of even numbers */
osum = 0;                        /* To store sum of odd numbers */

/* To generate the integer from 1 to n */

for(i=1; i<=n; i++)
if(i % 2 == 0)                     /* If even */
esum = esum + i;               /* Add to even sum */
osum = osum + i;


Thus, after the control comes out of the for-loop, the variable esum contains the sum of even numbers and the variable osum contains the sum of odd numbers. The result can be printed using the following statements:

printf("Sum of even numbers = %d\n",esum);
printf("Sum of odd numbers = %d\n", osum);

The complete program is shown below:

Example : Program to add even numbers and odd numbers.
main( )
int n, i, esum, osum;
printf("Enter the value of n:\n");
/* Generate and print number from 1 to n */
printf("Integer from 1 to %d are \n",n);
for(i=1; i<=n;i++)

/* find sum of even and odd numbers */
for ( i=1; i<=n; i++)
if((i %2) == 0)
esum +=i;     /* Add even numbers */
osum += i;    /* Add odd numbers */

printf(" Sum of even numbers = %d\n",esum);
printf("Sum of odd numbers = %d\n",osum);


Execution starts from here

Non-executable statement
Enter the value of n:
Output 1
Integer from 1 to 6
1 2 3 4 5 6

e_sum = 0
o_sum = 0

i=1 2 3 4 5 6
e_sum = 0+2+4+6
                         = 12
o_sum= 0+1+3+5
                         = 9
Output 2

Sum of even numbers =12
Sum of odd numbers = 9


Popular posts from this blog

difference between structure and union in C Language

In c language article we will see the difference between union and structure. Both are the user define datatype in c language. See the table which is mentioned below: ASP.NET Video Tutorial Series Structure Union1.The keywordstruct is used to define a structure 1. The keyword union is used to define a union. 2. When a variable is associated with a structure, the compiler allocates the memory for each member. The size of structure is greater than or equal to the sum ofsizes of its members. The smaller members may end with unused slack bytes. 2. When a variable is associated with a union, thecompiler allocates thememory by considering the size of the largest memory. So, size of union is equal to the size of largest member. 3. Each member within a structure is assigned unique storage area of location. 3. Memory allocated is shared by individual members of union. 4. The address of each member will be in ascending order This indicates that memory for each member will start at different offset v…

Difference between Linear search and Binary Search in c language

SQL Video Channel : Download all SQL Video

Binary Search Linear Search Works only on sorted items. such as  1,2,3,4,5,6  etc
Works on sorted as well as unsorted items. 12,4,5,3,2,1 etc Very efficient if the items are sorted Very efficient if the items are less and present in the beginning of the list. such as Suppose your list items are : 12,3,4,5,1 and you want to search 12 number then you get beginning in the list. Works well with arrays and not on linked lists. Works with arrays and linked lists.
Number of comparisons are less More number of comparisons are required if the items are present in the later part of the array or its elements are more.

Memory representation of Linked List Data Structures in C Language

Memory representation of Linked List

             In memory the linked list is stored in scattered cells (locations).The memory for each node is allocated dynamically means as and when required. So the Linked List can increase as per the user wish and the size is not fixed, it can vary.

               Suppose first node of linked list is allocated with an address 1008. Its graphical representation looks like the figure shown below:

      Suppose next node is allocated at an address 506, so the list becomes,

  Suppose next node is allocated with an address with an address 10,s the list become,

The other way to represent the linked list is as shown below:

 In the above representation the data stored in the linked list is “INDIA”, the information part of each node contains one character. The external pointer root points to first node’s address 1005. The link part of the node containing information I contains 1007, the address of next node. The last node …