Question

#include <iostream>

void swap(int &pi, int &pj){

    std::cout << "In function swap: " <<  &pi << " " << &pj << "\n";

        int temp = pi;
        pi = pj;
        pj = temp;
}

int main(){

        int i = 10, j = 20;
        int *pi = &i, *pj = &j;
        swap(pi, pj);

        std::cout << *pi << " " << *pj;
        return 0;
}

The above program does not give any compilation error. (Though to swap function in not POINTER TO REFERENCE type) and gives the proper output.
But whatever i am trying to print inside “swap” function is not printed to console.
Can anybody explain me why?

Answer 1

Looks like you’re probably using std::swap to swap two pointers instead of calling your own swap routine. I suspect you have a using namespace std; somewhere that you are not showing us ? Try changing the name of your swap routine to e.g. my_swap and then see if calling my_swap works (it should fail with a compilation error).