How do I make std::sort not have name collision between std::swap and my namespace's templated swap?

Question

I want to use std::sort, but the compile is failing with error C2668: std::swap: ambiguous call to overloaded function because there is a templated swap() function defined in my namespace that would be hard to get rid of. I don't care which swap it uses, but how do I make either one of them go away when compiling sort()?

I understand that it's ambiguous because my::swap is in the same namespace as my::Obj, and I don't care which version of swap gets used. I just have to overcome the namespace collision. This is part of a very large code base that I don't own so I'm hoping for a solution that is local to my code and presumably allows my::Obj and my::swapto both stay in namespace my.

namespace my
{
    template<class T> void swap(T a, T b)
    {
    }

    struct Obj
    {
    };

    void doSortStuff()
    {
        std::vector<Obj> arr;
        std::sort(arr.begin(), arr.end());
    }
};

Answer 1

A workaround is to create a better overload:

// No modifiable code
namespace my
{
    template<class T> void swap(T a, T b) { /*.. */ }
    struct Obj { /*..*/ };
}

// Your code:
namespace my
{
    void swap(Obj& lhs, Obj& rhs)
    {
        // my::swap<Obj&>(lhs, rhs);
        std::swap(lhs, rhs);
    }
}

// In namespace you want.
void doSortStuff()
{
    std::vector<my::Obj> arr;
    std::sort(arr.begin(), arr.end());
}

Then, between the 3 valid overloads, all are exact match, but the non-template is preferred.