Find all the even and odd numbers in an array

Posted by MindBreaker at 1:44 AM Read our previous post

Object:

4. Find all the even and odd numbers in an array

Source Code;

// EVEN ODD NUMBERS

#include <iostream>

using namespace std;

class average

{

public:

       int even,odd,count,elementeven[10],elementodd[10];

       int *sepeven( int ary[10])

       {

       for(count=0;count<10;count++)

       {

              if(ary[count]%2==0)

              {

                     even=ary[count];

                     elementeven[count]=even;

              }

      

       }

       return elementeven;

      

      

       }

       int *sepodd( int ary[10])

       {

       for(count=0;count<10;count++)

       {

              if(ary[count]%2!=0)

              {

                     odd=ary[count];

                     elementodd[count]=odd;

              }

      

             

      

       }

       return elementodd;

      

      

       }

};

void main()

{

       average avg;

       int i,arry[10],*retrneven,*retrnodd;

       cout<<"Enter numbers in array : \n";

      

       for(i=0;i<10;i++)

       {

      

       cin>>arry[i];

      

       }

       retrneven=avg.sepeven(arry);

       retrnodd=avg.sepodd(arry);

       for(i=0;i<10;i++)

       {cout<<"Even number are : "<<retrneven[i]<<endl;

      

       cout<<"Odd number are : "<<retrnodd[i]<<endl;

      

       }

}

      

      

  

 

  

}

Output: