Modify the function is_even in program below. The function should use pointer arithmetic not subscripting to visit array elements. In other words, eliminate the loop index variables and all use of the [] operator in the function.

void is_even(int * numbers, int n, int * result);

The is_even function determines if an element in the array is an even number. If it is even, the function stores 1 at the corresponding element in the result array, otherwise, it stores 0.

The main function remains the same as in the code below.

Sample run:

Enter the number of integers: 5

Enter the array elements: 32 24 7 21 19

Output: The output array: 1 1 0 0 0

This is the code to be modified below:

#include

void is_even(int numbers[], int n, int result[]);

//A program that determines if an element in an array is an even or odd number

int main()

{

//Declaring numbers and result as and integer array that can store up to 20 numbers.

//Declaring n and i as an integer.

int numbers[20];

int n;

int result[20];

int i;

//Enter the numbers of intergers for the array

printf("Enter the number of integers: ");

scanf("%d", &n);

//Enter the values to be saved in the array

printf("Enter the array elements: ");

//Using a loop statement to read in and display the outputs

for(i = 0; i < n; i++)

scanf("%d", &numbers[i]);

is_even(numbers, n, result);

printf("The output array: ");

for(i = 0; i < n; i++)

printf("%d ", result[i]);

printf(" ");

}

void is_even(int numbers[], int n, int result[])

{

//Using loop statement and a conditional statement to check if each value enter by the is even or odd.

//Displays 1 for even number and 0 for odd numbers.

int i;

for(i = 0; i < n; i++)

{

if(numbers[i] % 2 == 0) //is it even?

result[i] = 1;

else

result[i] = 0;

}

}