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; }