C program to count even and odd numbers in an array

C program to count even and odd numbers in an array

In this tutorial, we will discuss the concept of C program to count even and odd numbers in an array

In this post, we are going to learn how to count the even and odd numbers from the given array.

First, we must understand how to identify  odd or even numbers

What is Even and Odd

When the number is divided by 2 and the balance becomes zero and the above number is called as the even number – eg 2,4,6,8,10

and on the other sides when it is divided by 2 and balance becomes 1 they are called odd numbers or uneven numbers.

 

In my previous post, I have explained the various approaches to check whether the  number is even or odd in C language

here, we will discuss how to count odd and even nunbers from an array in  the C programming language using loops

 

There are three programs given below

The common procedure of this programs

• Initially, Declare an integer type array with maximum capacity(100) and declare variables i, size, odd and even
• The user will enter the size of the array and store in the variable size to define the array size
• Then, the user enters all the elements of the array according to the array size using loops
• After that, use a loop for iterating elements of the array
• Finally, the following  loop count even and odd number using if statements from the array

Count even and odd numbers in an array using for loop

Program 1

The program allows the user to enter the size of the array and receives the elements from the user.

Then, it will count the even and odd numbers from this array using for loop in C

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int arr[100];
    int i,size,odd=0,even=0;

    //input the size of the array
    printf("Enter size of the array\n");
    scanf("%d",&size);//reads input from user for array size

    //input the array elements
    printf("\nEnter elements of the array\n\n");
    for(i=0; i<size; i++)
    {
        printf("Enter the element arr[%d] :",i);
        scanf("%d",&arr[i]);//reads input from user for array elements
    }


    for(i=0; i<size; i++)
    {
        if(arr[i]%2==0)
        {
            even++;
        }
        else{
            odd++;
        }
    }

    printf("\nTotal even numbers of an array :%d\n",even);//display even numbers
     printf("Total odd numbers of an array :%d",odd);//display odd numbers
     getch();
    return 0;
}

When the above code is executed, it produces the following results

Enter the size of the array
10
Enter elements of the array

Enter the element arr[0] : 45
Enter the element arr[1] : 65
Enter the element arr[2] : 68
Enter the element arr[3] : 87
Enter the element arr[4] : 98
Enter the element arr[5] : 12
Enter the element arr[6] : 54
Enter the element arr[7] : 66
Enter the element arr[8] : 67
Enter the element arr[9] : 42

Total even numbers of an array :6
Total odd numbers of an array : 4

 

Count even and odd numbers in an array using while loop

Program 1

The program allows the user to enter the size of the array and receives the elements from the user.

Then, it will count the even and odd numbers from this array using while loop in C

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int arr[100];
    int i,size,odd=0,even=0;

    //input the size of the array
    printf("Enter size of the array\n");
    scanf("%d",&size);//reads input from user for array size

    //input the array elements
    printf("\nEnter elements of the array\n\n");
    i=0;
    while(i<size)
    {
        printf("Enter the element arr[%d] :",i);
        scanf("%d",&arr[i]);//reads input from user for array elements
         i++;
    }


    i=0;
    while(i<size)
    {
        if(arr[i]%2==0)
        {
            even++;
        }
        else{
            odd++;
        }
         i++;
    }

    printf("\nTotal even numbers of an array :%d\n",even);//display even numbers
     printf("Total odd numbers of an array :%d",odd);//display odd numbers
     getch();
    return 0;
}

 

When the above code is executed, it produces the following results

Enter the size of the array
8
Enter elements of the array

Enter the element arr[0] : 987
Enter the element arr[1] : 876
Enter the element arr[2] : 765
Enter the element arr[3] : 654
Enter the element arr[4] : 456
Enter the element arr[5] : 653
Enter the element arr[6] : 542
Enter the element arr[7] : 457


Total even numbers of an array :4
Total odd numbers of an array : 4

 

Count even and odd numbers in an array using the do-while loop

Program 1

The program allows the user to enter the size of the array and receives the elements from the user.

Then, it will count the even and odd numbers from this array using the do-while loop in C

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int arr[100];
    int i,size,odd=0,even=0;

    //input the size of the array
    printf("Enter size of the array\n");
    scanf("%d",&size);

    //input the array elements
    printf("\nEnter elements of the array\n\n");
    i=0;

    do{
        printf("Enter the element arr[%d] :",i);
        scanf("%d",&arr[i]);
         i++;
    }while(i<size);


    i=0;

    do{
        if(arr[i]%2==0)
        {
            even++;
        }
        else{
            odd++;
        }
         i++;
    }while(i<size);

    printf("\nTotal even numbers of an array :%d\n",even);//display even numbers
     printf("Total odd numbers of an array :%d",odd);//display odd numbers
     getch();
    return 0;
}

 

When the above code is executed, it produces the following results

Enter the size of the array
8
Enter elements of the array

Enter the element arr[0] : 900
Enter the element arr[1] : 855
Enter the element arr[2] : 254
Enter the element arr[3] : 545
Enter the element arr[4] : 124
Enter the element arr[5] : 578
Enter the element arr[6] : 980
Enter the element arr[7] : 642


Total even numbers of an array :6
Total odd numbers of an array : 2

 

Similar post

Java code to count even and odd numbers of an array

C++ code to count even and odd numbers of an array

 

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