Write a function that removes all even numbers from an array

Write a function that removes all even numbers from an array. The function should take the array, length of the array, and a pointer for number of odd numbers found as arguments and return a pointer to the new array. If there are no odd numbers found in the array, your code should print \"No odds found.\" and return NULL Use the function header: int removeEvens(int a, int length, int oddsFound); Example: Input array 13, 4, 5, 6, 7) *oddsFound 3 return array 13, 5, 7) The size of the return array needs to be the number of odd numbers found. Note: You c

Solution

ANSWER TO FIRST QUESTION

#include<iostream.h>
#include<conio.h>
int* removeEvens(int *a,int length, int *oddsFound)

{
     int *odd_arr;
     odd_arr=new int[*oddsFound];
     int c=0;
     for(int i=0;i<length;i++)
     {
   if(a[i]%2!=0)
   {
       odd_arr[c]=a[i];
       c++;
   }
     }
     return odd_arr;

}
void main()
{
   int odd=0;
   int a[50];
   int n;
   int *p;
   clrscr();

   cout<<\"enter the number of elements\"<<endl;
   cin >> n;
   for(int i=0;i<n;i++)
   cin >> a[i];

   for(int j=0;j<n;j++)
   {
       if(a[j]%2!=0)
       {
           odd=odd+1;
       }
   }

       p=removeEvens(a,n,&odd);

       cout<<\"the content of odd array are\"<<endl;
       for(int k=0;k<odd;k++)
       cout<<\"\ \"<<p[k];

   getch();

}

ANSWER TO THE SECOND QUESTION

option c is the answer. 10 8 10 8 10 will be the contents of the array. after executing that code.

answer to third question

option d is not the valid function call.