Confusion about pointers and references in C++

[Harshjain]

I have a bunch of code like this:

#include <iostream>
using namespace std;

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

int main() {
    int a;
    int b;
    a = 7;
    b = 5;
    swap(a, b);
    cout << a << b;

    return 0;
}

This code does the swapping process as what I precisely needed to swap 2 numbers, however when I need two numbers from the user as follows;

int a;
int b;
cin >> a;
cin >> b;
swap(a, b);
cout << a <<

the compiler gives me an error about int to int* error which is as expected. Why does the first code do the right swapping although I didn't use the method with & operator?

[2kaud]

Interesting question. The real question is why the first code compiles. The given swap() function takes 2 args of pointers to ints. However you are passing ints in the function call. So why does this compile and work?

The answer is in the using statement. C++ has a std::swap() function. With the using statement (and iostream also using utility) then the compiler matches the swap() call to the C++ standard one and not the provided one.

Try this - and you should get the error:

#include <iostream>
//using namespace std;

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

int main() {
	int a;
	int b;
	a = 7;
	b = 5;
	swap(a, b);
	std::cout << a << b;

	return 0;
}

The moral is - If you use using namespace std, don't have functions with the same name as those used by C++.

This is OK:

#include <iostream>
//using namespace std;

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

int main() {
	int a;
	int b;
	a = 7;
	b = 5;
	swap(&a, &b);
	std::cout << a << b;

	return 0;
}

or just:

#include <iostream>
//using namespace std;

int main() {
	int a;
	int b;
	a = 7;
	b = 5;
	std::swap(a, b);
	std::cout << a << b;

	return 0;
}

Note that you should have #include <utility> instead if relying upon this being provided by other includes.

See https://cplusplus.com/reference/utility/swap/