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



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

Count Even/Odd Numbers in Array of 10 Numbers

The question is, write a program in C++ that receives an array of 10 numbers and count even/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 above C++ program on counting 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 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 above program goes like:

Count Even/Odd Numbers in Array of n Numbers

This is the modified version of previous program, because this program allows user to define the size of 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, 7 as four array elements:

c++ program count even odd numbers array

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