C program to count even and odd elements in array

Write a C program to input elements in array from user and count even and odd elements in array. How to find total number of even and odd elements in a given array using C programming. Logic to count even and odd elements in array using loops.

Input array: 1 2 3 4 5 6 7 8 9

Total even elements: 4
Total odd elements: 5

Required knowledge

Basic Input Output, If else, Loop, Array

Logic to count even and odd elements in array

Step by step descriptive logic to count total even and odd elements in array.

1. Input size and elements in array from user. Store it in some variable say size and arr.
2. Declare and initialize two variables with zero to store even and odd count. Say even = 0 and odd = 0.
3. Iterate through each array element. Run a loop from 0 to size - 1. Loop structure should look like for(i=0; i<size; i++).
4. Inside loop increment even count by 1 if current array element is even. Otherwise increment the odd count.
   

   Learn different ways of checking even odd.

   • Program to check even odd using if else.
   • Program to check or odd using conditional statement.
   • Program to check even odd using switch case.
   • Program to check even odd using bitwise operator.

5. Print the values of even and odd count after the termination of loop.

Program to count even and odd elements in array

/**
 * C program to count total number of even and odd elements in an array
 */

#include <stdio.h>

#define MAX_SIZE 100 //Maximum size of the array

int main()
{
    int arr[MAX_SIZE];
    int i, size, even, odd;

    /* Input size of the array */
    printf("Enter size of the array: ");
    scanf("%d", &size);

    /* Input array elements */
    printf("Enter %d elements in array: ", size);
    for(i=0; i<size; i++)
    {
        scanf("%d", &arr[i]);
    }

    /* Assuming that there are 0 even and odd elements */
    even = 0;
    odd  = 0;

    for(i=0; i<size; i++)
    {
        /* If the current element of array is even then increment even count */
        if(arr[i]%2 == 0)
        {
            even++;
        }
        else
        {
            odd++;
        }
    }

    printf("Total even elements: %d\n", even);
    printf("Total odd elements: %d", odd);

    return 0;
}

Enter size of the array: 10
Enter 10 elements in array: 5 6 4 12 19 121 1 7 9 63
Total even elements: 3
Total odd elements: 7

Happy coding 😉