Specializing a function for a private class?

Question

Is there any way to specialize a function (say, std::swap) for a private class?

For example, when I test this:

#include <algorithm>

class Outer
{
    struct Inner
    {
        int a;
        void swap(Inner &other)
        {
            using std::swap;
            swap(this->a, other.a);
        }
    };
public:
    static void test();
};

namespace std
{
    template<> void swap<Outer::Inner>(Outer::Inner &a, Outer::Inner &b)
    { a.swap(b); }
}
void Outer::test()
{
    using std::swap;
    Inner a, b;
    swap(a, b);
}
int main()
{
    Outer::test();
    return 0;
}

I get this:

Test.cpp:20:47: error: 'Inner' is a private member of 'Outer'
    template<> void swap<Outer::Inner>(Outer::Inner &a, Outer::Inner &b)
                                              ^
Test.cpp:5:12: note: implicitly declared private here
    struct Inner
           ^
Test.cpp:20:64: error: 'Inner' is a private member of 'Outer'
    template<> void swap<Outer::Inner>(Outer::Inner &a, Outer::Inner &b)
                                                               ^
Test.cpp:5:12: note: implicitly declared private here
    struct Inner
           ^
Test.cpp:20:33: error: 'Inner' is a private member of 'Outer'
    template<> void swap<Outer::Inner>(Outer::Inner &a, Outer::Inner &b)
                                ^
Test.cpp:5:12: note: implicitly declared private here
    struct Inner

(I do realize declaring a friend swap that can be found through ADL avoids this issue for swap, but that's irrelevant to my question. swap is just an example.)

Answer 1

You could add a friend declaration of the std::swap<Inner>(Inner&, Inner&) inside Outer

#include <algorithm>

class Outer
{
    struct Inner
    {
        int a;
        void swap(Inner &other)
        {
            using std::swap;
            swap(this->a, other.a);
        }
    };

    friend void std::swap<Inner>(Inner&, Inner&) noexcept;
public:
    static void test();
};

namespace std
{
    template<> void swap<Outer::Inner>(Outer::Inner &a, Outer::Inner &b) noexcept
    { a.swap(b); }
}

void Outer::test()
{
    using std::swap;
    Inner a, b;
    swap(a, b);
}

int main()
{
    Outer::test();
    return 0;
}

Live Example

Answer 2

Don't extend the std namespace.

If you want to create a swap function for Inner, make it a private function in Outer

#include <algorithm>

class Outer
{
    struct Inner
    {
        int a;
        void swap(Inner &other)
        {
            std::swap(this->a, other.a);
        }
    };

    static void swap(Inner& a, Inner& b);

public:
    static void test();
};

void Outer::test()
{
    Inner a, b;
    swap(a, b);
}

void Outer::swap(Inner& a, Inner& b)
{
    a.swap(b);
}

int main()
{
    Outer::test();
    return 0;
}