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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
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