Array Exercises
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.
Required knowledge
Basic C programming, If else, c-Arrays, C Loops
Logic to count even and odd elements in array
Step by step descriptive logic to count total even and odd elements in array.
- Input size and elements in array from user. Store it in some variable say size and arr.
- Declare and initialize two variables with zero to store even and odd count. Say even = 0 and odd = 0.
- Iterate through each array element. Run a loop from 0 to size - 1. Loop structure should look like for(i=0; i<size; i++).
- Inside loop increment even count by 1 if current array element is even. Otherwise increment the odd count.
- 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;
}
