How to overload std::swap()

Question

std::swap() is used by many std containers (such as std::list and std::vector) during sorting and even assignment.

But the std

Answer 1

Attention Mozza314

Here is a simulation of the effects of a generic std::algorithm calling std::swap, and having the user provide their swap in namespace std. As this is an experiment, this simulation uses namespace exp instead of namespace std.

// simulate 

#include 

namespace exp
{

    template 
    void
    swap(T& x, T& y)
    {
        printf("generic exp::swap\n");
        T tmp = x;
        x = y;
        y = tmp;
    }

    template 
    void algorithm(T* begin, T* end)
    {
        if (end-begin >= 2)
            exp::swap(begin[0], begin[1]);
    }

}

// simulate user code which includes 

struct A
{
};

namespace exp
{
    void swap(A&, A&)
    {
        printf("exp::swap(A, A)\n");
    }

}

// exercise simulation

int main()
{
    A a[2];
    exp::algorithm(a, a+2);
}

For me this prints out:

generic exp::swap

If your compiler prints out something different then it is not correctly implementing "two-phase lookup" for templates.

If your compiler is conforming (to any of C++98/03/11), then it will give the same output I show. And in that case exactly what you fear will happen, does happen. And putting your swap into namespace std (exp) did not stop it from happening.

Dave and I are both committee members and have been working this area of the standard for a decade (and not always in agreement with each other). But this issue has been settled for a long time, and we both agree on how it has been settled. Disregard Dave's expert opinion/answer in this area at your own peril.

This issue came to light after C++98 was published. Starting about 2001 Dave and I began to work this area. And this is the modern solution:

// simulate 

#include 

namespace exp
{

    template 
    void
    swap(T& x, T& y)
    {
        printf("generic exp::swap\n");
        T tmp = x;
        x = y;
        y = tmp;
    }

    template 
    void algorithm(T* begin, T* end)
    {
        if (end-begin >= 2)
            swap(begin[0], begin[1]);
    }

}

// simulate user code which includes 

struct A
{
};

void swap(A&, A&)
{
    printf("swap(A, A)\n");
}

// exercise simulation

int main()
{
    A a[2];
    exp::algorithm(a, a+2);
}

Output is:

swap(A, A)

Update

An observation has been made that:

namespace exp
{    
    template <>
    void swap(A&, A&)
    {
        printf("exp::swap(A, A)\n");
    }

}

works! So why not use that?

Consider the case that your A is a class template:

// simulate user code which includes 

template 
struct A
{
};

namespace exp
{

    template 
    void swap(A&, A&)
    {
        printf("exp::swap(A, A)\n");
    }

}

// exercise simulation

int main()
{
    A a[2];
    exp::algorithm(a, a+2);
}

Now it doesn't work again. :-(

So you could put swap in namespace std and have it work. But you'll need to remember to put swap in A's namespace for the case when you have a template: A. And since both cases will work if you put swap in A's namespace, it is just easier to remember (and to teach others) to just do it that one way.