Home >Backend Development >C++ >How to separate even and odd numbers in an array using for loop in C language?
An array is a group of related data items stored under a single name.
For example int Student[30]; //student is an array name, a collection of 30 data items containing a single variable name
Search - Used to find if a particular element is present
Sort - It helps in arranging the elements in an array in ascending or descending order arrangement.
Traversal - It processes each element in the array sequentially.
Insertion - It helps to insert elements in an array.
Delete - It helps in deleting elements from the array. elements in the array.
The logic of finding even numbers in the array is as follows-
for(i = 0; i < size; i ++){ if(a[i] % 2 == 0){ even[Ecount] = a[i]; Ecount++; } }
The logic of finding odd numbers in the array is as follows-
for(i = 0; i < size; i ++){ if(a[i] % 2 != 0){ odd[Ocount] = a[i]; Ocount++; } }
To display even numbers, please call the display function mentioned below -
printf("no: of elements comes under even are = %d </p><p>", 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("</p><p>"); }
To display odd numbers, please call the display function as follows −
printf("no: of elements comes under odd are = %d </p><p>", 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("</p><p>"); }
The following is a C program that uses for loop to separate even numbers and odd numbers in an array-
Live demonstration
#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 :</p><p>"); scanf("%d", &size); printf("enter array elements:</p><p>"); 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 </p><p>", Ecount); printf("The elements that are present in an even array is: "); display(even, Ecount); printf("no: of elements comes under odd are = %d </p><p>", 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("</p><p>"); }
When the above program is executed, the following results will be produced-
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
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