C++ Program to Count Even and Odd Numbers in an Array

This article provides you with some programs in C++ that count the number of even and odd numbers available in an array. The array must be entered by the user at run-time.

Count the even and odd numbers in a 10-number array

The question is: write a program in C++ that receives an array of 10 numbers and counts the even and odd numbers available in the given array. The program given below is the answer to this question:

#include<iostream>

using namespace std;
int main()
{
   int arr[10], eve=0, odd=0, i;
   cout<<"Enter 10 Array Elements: ";
   for(i=0; i<10; i++)
      cin>>arr[i];
   for(i=0; i<10; i++)
   {
      if(arr[i]%2==0)
         eve++;
      else
         odd++;
   }
   cout<<"\nTotal Number of Even Numbers = "<<eve;
   cout<<"\nTotal Number of Odd Numbers = "<<odd;
   cout<<endl;
   return 0;
}

The snapshot given below shows the initial output produced by the above C++ program on counting the total number of even and odd numbers in a given array:

c++ count even odd numbers in array

Now supply the input, say 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 as ten array elements, and press the ENTER key to find and print the number of odd and even numbers like shown in the snapshot given below:

count odd even numbers in array c++

The dry run of the above program goes like this:

Count the number of even and odd numbers in an array of n numbers

This is the modified version of the previous program because this program allows the user to define the size of the array along with its elements.

#include<iostream>

using namespace std;
int main()
{
   int tot, arr[100], i, eve=0, odd=0;
   cout<<"Enter the Size of Array (max. 100): ";
   cin>>tot;
   cout<<"Enter "<<tot<<" Array Elements: ";
   for(i=0; i<tot; i++)
      cin>>arr[i];
   for(i=0; i<tot; i++)
   {
      if(arr[i]%2==0)
         eve++;
      else
         odd++;
   }
   cout<<endl;
   if(eve>1)
      cout<<"There are "<<eve<<" Even Numbers.";
   else
   {
      if(eve==1)
         cout<<"There is only 1 Even Number.";
      else
         cout<<"There is no any Even Number.";
   }
   cout<<endl;
   if(odd>1)
      cout<<"There are "<<odd<<" Odd Numbers.";
   else
   {
      if(odd==1)
         cout<<"There is only 1 Odd Number.";
      else
         cout<<"There is no any Odd Number.";
   }
   cout<<endl;
   return 0;
}

Here is its sample run with user input, 4 as size, and 1, 2, 5, and 7 as four array elements:

c++ program count even odd numbers array

C++ Quiz


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