Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
C program to find array type entered by user using pointers.
Problem
Write a C Program to find the array type that we need to check, whether the given elements in an array are even numbers or odd numbers or combination of both by using pointers.
Solution
The user has to enter an array of integers, then, display the type of the array.
Example 1 − Input: 5 3 1, Output: odd array
Example 2 − Input: 2 4 6 8, Output: even array
Example 3 − Input: 1 2 3 4 5, Output: mixed array
Algorithm
Refer an algorithm given below to find the array type entered by user with the help of pointers.
Step 1: Read the size of array at runtime.
Step 2: Input the array elements.
Step 3: Declare the pointer variable.
Step 3: Check, if all the elements of the array are odd by using pointer variable.
Then, print "Odd".
Step 4: Check, if all the elements of the array are even by using pointer variable.
Then, print "Even".
Step 5: Else, print "Mixed".
Example
Following is the C program to find the array type entered by user with the help of pointers −
#include<stdio.h>
#include<stdlib.h>
int*createArray (int);
void readArray(int,int *);
int findType(int , int *);
int main(){
int *a,n,c=0,d=0;
printf("Enter the size of array
");
scanf("%d",&n);
printf("Enter the elements of array
");
createArray(n);
readArray(n,a);
findType(n,a);
return 0;
}
int *createArray(int n){
int *a;
a=(int*)malloc(n*sizeof(int));
return a;
}
void readArray(int n,int *a){
for(int i=0;i<n;i++){
scanf("%d",a+i);
}}
int findType(int n, int *a){
int c=0,d=0;
for(int i=0;i<n;i++){
if(a[i]%2==0){
c++;
}
else{
d++;
}}
if(c==n){
printf("The array type is Even
");
}
if(d==n){
printf("The array type is Odd
");
}
if(c!=n && d!=n){
printf("The array type is Mixed
");
}
return 0;
}Output
When the above program is executed, it produces the following output −
Enter the size of array 4 Enter the elements of array 12 14 16 18 The array type is Even
447 Views
