I need some help with this question

Question

Complete the missing parts of this program. The inserted comments will help you in specifying the functionality of this program:

#include<iostream >
using namespace std;
void exchange(int *a, int index1, int index2)
{
​//This function swaps the two elements (specified by the parameters index1
//and index2) of the passed array (a).
// Note: Do not use array notation ([]). You can only use pointer notations.

}

void printElements (int *a,int size)
{
​// This function prints the array elements in one line separated by 3 spaces.
// Note: Do not use array notation ([]). You can only use pointer notations.

}

void reverseElements(int *a,int size)
{
​// This function stores the array elements in reverse order.
// Note: Do not use array notation ([]). You can only use pointer notations.
​

}
void main()
{
​int a[5] = {10,20,30,40};
​
​exchange(a,2,3);
​
​printElements(a,5);

​reverseElements(a,5);

​printElements(a,5);​
​
}

Answer 1

Array notation is a[index1] and pointer notation for this equalent is *(a+index1);

By this you can implement swap and print elements..
For swap use a temporary variable and to print use a for loop from index 0 to 5..
For reverse, just traverse array elements from 5 to 0.

Try these, and post your try if any problem...
Hope it helps.

Answer 2

#include <iostream>
using namespace std;
void exchange(int *a, int index1, int index2)
{
*a = index1;
index1 = index2;
index2 = *a;
// to swaps the two elements
}

void printElements(int *a, int size)
{

for (; *a != '\0'; a++)
cout << *a << " ";
// to elements in one line
}

void reverseElements(int *a, int size)
{

int *p1 = a; // *p1 pointing at the beginning of the array

int *p2 = a + size - 1; // p2 pointing at end of the array
while (p1 < p2) {
swap(p1, p2);
p1++;
p2--;
}

}
int main() {
int a[5] = { 10,20,30,40 };

exchange(a, 2, 3);

printElements(a, 5);

reverseElements(a, 5);

printElements(a, 5);

return 0;
}
sorry for being late
is this right ??