Question

The C++ standard prohibits declaring types or defining anything in namespace std, but it does allow you to specialize standard STL templates for user-defined types.

Usually, when I want to specialize std::swap for my own custom templated type, I just do:

namespace std
{
  template <class T>
  void swap(MyType<T>& t1, MyType<T>& t2)
  {
     t1.swap(t2);
  }
}

…and that works out fine. But I’m not entirely sure if my usual practice is standard compliant. Am I doing this correctly?

Answer 1

What you have is not a specialization, it is overloading and exactly what the standard prohibits. (However, it will almost always currently work in practice, and may be acceptable to you.)

Here is how you provide your own swap for your class template:

template<class T>
struct Ex {
  friend void swap(Ex& a, Ex& b) {
    using std::swap;
    swap(a.n, b.n);
  }
  T n;
}

And here is how you call swap, which you’ll notice is used in Ex’s swap too:

void f() {
  using std::swap; // std::swap is the default or fallback
  Ex<int> a, b;
  swap(a, b); // invokes ADL
}

Related: Function template specialization importance and necessity