Confusion about pointers and references in C++

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 exactly wanted to swap 2 numbers

But when I want two numbers from the user as follows;

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

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

---

In the first example, std::swap is called, because of your using namespace std . The second example is exactly the same as the first one, so you might have no using.

Anyway, if you rename your function to my_swap or something like that (and change every occurence), then the first code shouldn't work, as expected. Or, remove the using namespace std and call std::cin and std::cout explicitly. I would recommend the second option.

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