c++ Swap function declared with a pointer parameter can be called with a value as a parameter but not an address? [duplicate]

I'm beginning with c++, I'm writing a program to sort array with int data type, with swap function as follows:

void swap(int *a, int *b){
    int m;
    m = *a;
    *a = *b;
    *b = m;
}

I know when calling swap function we have to pass parameter to it as an address (ex: swap(&a[i], &a[j]) but if I try passing parameter as a value (ex : swap(a[i], a[j]), swap function is working properly, I'm a bit confused about this, an explanation would help me a lot, many thanks

Sorry, I edited, here is my full source code

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<conio.h>
using namespace std;

void swap(int *a, int *b){
    int m;
    m = *a;
    *a = *b;
    *b = m;
}
void bubblesort(int * a, int n){
    for (int i = 0; i < n - 1; i++){
        for (int j = i + 1; j < n; j++){
            if (a[i] <= a[j]) swap(a[i], a[j]);
        }
    }
}
int main(){
    int aray[6] = {2,4,6,1,9,7};
    bubblesort(aray, 6);

    for (int i = 0; i < 6; i++)
    {
        cout << aray[i];
    }

    _getch();
    return 0;
}