C++ Program to Print all Elements available on Even Positions

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This article provides some programs in C++ that find and prints elements or numbers available on even position in an array. Here are list of programs provided here:

Print elements on even position (index-wise)
Print elements on even position (normal position-wise)

Print Elements on Even Position - Index-wise

The question is, write a C++ program that receives 10 elements for the array to find and print all elements available on even positions. Here is its answer:

#include<iostream>

using namespace std;
int main()
{
   int arr[10], i;
   cout<<"Enter 10 array elements: ";
   for(i=0; i<10; i++)
      cin>>arr[i];
   cout<<"\nList of elements available on even positions:\n";
   for(i=0; i<10; i++)
   {
      if(i%2==0)
         cout<<arr[i]<<" ";
   }
   cout<<endl;
   return 0;
}

The snapshot given below shows the initial output produced by above C++ program on printing all even positioned elements:

c++ print elements on even positions array

Now supply the input say 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 as ten numbers and press ENTER key to find and print all elements or numbers available on even indexes like shown in the snapshot given below:

The above program works only on 10 elements. That is, to allow user to work with required number of elements, then here is another program created after modifying the above one:

#include<iostream>

using namespace std;
int main()
{
   int arr[100], i, n;
   cout<<"How many elements (numbers) to enter ? ";
   cin>>n;
   cout<<"Enter "<<n<<" array elements: ";
   for(i=0; i<n; i++)
      cin>>arr[i];
   cout<<"\nList of elements available on even positions:\n";
   for(i=0; i<n; i++)
   {
      if(i%2==0)
         cout<<arr[i]<<" ";
   }
   cout<<endl;
   return 0;
}

Here is its sample run with user input 4 as size and 1, 2, 3, 4 as its elements:

print element on even position c++ program

Note - Since the maximum size defined for the array arr in above program is 100. Therefore, don't provide the size more than this number. But I've created another program, that defines the size of array arr dynamically, as shown in the upcoming program.

Print Elements on Even Index using Second Array

This program uses another array to store all the elements available on even index numbers one by one.

#include<iostream>

using namespace std;
int main()
{
   int n, i, j=0;
   cout<<"How many elements (numbers) to enter ? ";
   cin>>n;
   int arr[n], eve[n];
   cout<<"Enter "<<n<<" array elements: ";
   for(i=0; i<n; i++)
   {
      cin>>arr[i];
      if(i%2==0)
      {
         eve[j] = arr[i];
         j++;
      }
   }
   cout<<"\nList of elements available on even positions:\n";
   for(i=0; i<j; i++)
      cout<<eve[i]<<" ";
   cout<<endl;
   return 0;
}

This program produces exactly same output as of previous program.

Print Element on Even Position - Normal Position-wise

This is the last program of this article that find and prints all the elements available on even position.

#include<iostream>

using namespace std;
int main()
{
   int n, i, j=0;
   cout<<"How many elements (numbers) to enter ? ";
   cin>>n;
   int arr[n], eve[n];
   cout<<"Enter "<<n<<" array elements: ";
   for(i=0; i<n; i++)
   {
      cin>>arr[i];
      if((i+1)%2==0)
      {
         eve[j] = arr[i];
         j++;
      }
   }
   cout<<"\nList of elements available on even positions:\n";
   for(i=0; i<j; i++)
      cout<<eve[i]<<" ";
   cout<<endl;
   return 0;
}

Here is its sample run with user input 10 as size and again 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 as ten array elements:

Note - The only change I've done in above program, in comparison with previous is, by replacing the following code:

if(i%2==0)

with the code given below:

if((i+1)%2==0)

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