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How to separate even and odd numbers in an array by using for loop in C language?
An array is a group of related data items that are stored with single name.
For example, int student[30]; //student is an array name that holds 30 collection of data items with a single variable name
Operations of array
Searching − It is used to find whether particular element is present or not
Sorting − It helps in arranging the elements in an array either in ascending or descending order.
Traversing − It processes every element in an array sequentially.
Inserting − It helps in inserting the elements in an array.
Deleting − It helps in deleting an element in an array.
The logic to find even numbers in an array is as follows −
for(i = 0; i < size; i ++){ if(a[i] % 2 == 0){ even[Ecount] = a[i]; Ecount++; } }
The logic to find odd numbers in an array is as follows −
for(i = 0; i < size; i ++){ if(a[i] % 2 != 0){ odd[Ocount] = a[i]; Ocount++; } }
To display even numbers, call display function as mentioned below −
printf("no: of elements comes under even are = %d
", Ecount); printf("The elements that are present in an even array is: "); void display(int a[], int size){ int i; for(i = 0; i < size; i++){ printf("%d \t ", a[i]); } printf("
"); }
To display odd numbers, call display function as given below −
printf("no: of elements comes under odd are = %d
", Ocount); printf("The elements that are present in an odd array is : "); void display(int a[], int size){ int i; for(i = 0; i < size; i++){ printf("%d \t ", a[i]); } printf("
"); }
Program
Following is the C program to separate even and odd numbers in an array by using for loop −
#include<stdio.h> void display(int a[], int size); int main(){ int size, i, a[10], even[20], odd[20]; int Ecount = 0, Ocount = 0; printf("enter size of array :
"); scanf("%d", &size); printf("enter array elements:
"); for(i = 0; i < size; i++){ scanf("%d", &a[i]); } for(i = 0; i < size; i ++){ if(a[i] % 2 == 0){ even[Ecount] = a[i]; Ecount++; } else{ odd[Ocount] = a[i]; Ocount++; } } printf("no: of elements comes under even are = %d
", Ecount); printf("The elements that are present in an even array is: "); display(even, Ecount); printf("no: of elements comes under odd are = %d
", Ocount); printf("The elements that are present in an odd array is : "); display(odd, Ocount); return 0; } void display(int a[], int size){ int i; for(i = 0; i < size; i++){ printf("%d \t ", a[i]); } printf("
"); }
Output
When the above program is executed, it produces the following result −
enter size of array: 5 enter array elements: 23 45 67 12 34 no: of elements comes under even are = 2 The elements that are present in an even array is: 12 34 no: of elements comes under odd are = 3 The elements that are present in an odd array is : 23 45 67