Why does passing int into function as pointer argument work when using namespace std is declared?

Question

I was messing with a function that swaps the value of two variables using pointer and I found out that by declaring using namespace std, I wouldn't need to pass the address of the variables as arguments for it to work. Passing the int itself would make the code works.

I suspect it has something to do with how std process arguments but I couldn't find an answer for it. I'd appreciate it if someone could point me to the right direction.

1) Without using namespace std

#include<iostream>
void swap(int* X, int* Y);
int main() {
    int a, b;
    std::cin >> a;
    std::cin >> b;
    std::cout << a << "," << b << std::endl;
    swap(&a, &b);
    std::cout << a << "," << b << std::endl;
}
void swap(int* X, int* Y) {
    int temp = *X;
    *X = *Y;
    *Y = temp;
}

2) With using namespace std

#include<iostream>
**using namespace std;**
void swap(int* X, int* Y);
int main() {
    int a, b;
    std::cin >> a;
    std::cin >> b;
    std::cout << a << "," << b << std::endl;
    swap(a, b);
    std::cout << a << "," << b << std::endl;
}
void swap(int* X, int* Y) {
    int temp = *X;
    *X = *Y;
    *Y = temp;
}

Answer 1

It doesn't. With using namespace std; you wind up calling std::swap which takes its parameters by reference. This is one of the many reasons that using namespace std is very bad practice.