Why does an array of pointers initializes itself when function finishes? - c++

Question

My weekend assignment was to make a function that gets an array of integers and the size of the array, and creates an array of pointers so that the pointers will be sorted using bubble sort (without changing the original array).
While debugging I found out that it works just fine, but when the function goes back to main() the pointers array gets initialized and everything's gone.
#include <iostream>
using namespace std;
void pointerSort(int arr[], int size, int* pointers[]);
void swap(int a, int b);
void main()
{
int arr[5]={7,2,5,9,4};
int size = 5;
int* pointers[5];
pointerSort(arr, size, pointers);
for (int i = 0; i < 5 ; i++)
cout << *pointers[i] << endl;
}
void pointerSort(int arr[], int size, int* pointers[])
{
int j, i;
bool change = true;
pointers = new int*[size];
for (i = 0; i < size; i++)
pointers[i] = &arr[i];
i = 0;
j = 1;
while (i <= size-1 && change == true)
{
change = false;
for (i = 0; i < size-j; i++)
{
if (*pointers[i] > *pointers[i+1])
{
swap(pointers[i], pointers[i+1]);
change = true;
}
}
j++;
}
}
void swap(int&a, int&b)
{
int temp;
temp = a;
a = b;
b = temp;
}

Answer 1

pointers = new int*[size];
At this point pointers is already an array of pointers, no allocation is needed.
After this line pointers IS NO LONGER THE ARRAY IN YOUR MAIN FUNCTION.
This is why your function is failing, because you are reassigning the array to which pointers is pointing to. The original array ISNT getting reinitialized, its just ignored throughout the entire code.
It is also a memory leak as ATaylor mentions, since you do not delete the allocated space, and cannot delete the space after the function finishes.
To fix everything: just remove the above line.