how to provide a swap function for my class?

What is the proper way to enable my swap in STL algorithms?

1) Member swap . Does std::swap use SFINAE trick to use the member swap .

2) Free standing swap in the same namespace.

3) Partial specialization of std::swap .

4) All of the above.

Thank you.

EDIT: Looks like I didn't word my question clearly. Basically, I have a template class and I need STL algos to use the (efficient) swap method I wrote for that class.

---

1) is the proper use of swap . Write it this way when you write "library" code and want to enable ADL (argument-dependent lookup) on swap . Also, this has nothing to do with SFINAE.

// some algorithm in your code
template<class T>
void foo(T& lhs, T& rhs){
  using std::swap; // enable 'std::swap' to be found
                   // if no other 'swap' is found through ADL
  // some code ...
  swap(lhs, rhs); // unqualified call, uses ADL and finds a fitting 'swap'
                  // or falls back on 'std::swap'
  // more code ...
}

---

2) Is the proper way to provide a swap function for your class.

namespace Foo{

class Bar{}; // dummy

void swap(Bar& lhs, Bar& rhs){
  // ...
}

}

If swap is now used as shown in 1), your function will be found. Also, you may make that function a friend if you absolutely need to, or provide a member swap that is called by the free function:

// version 1
class Bar{
public:
  friend void swap(Bar& lhs, Bar& rhs){
    // ....
  }
};

// version 2
class Bar{
public:
  void swap(Bar& other){
    // ...
  }
};

void swap(Bar& lhs, Bar& rhs){
  lhs.swap(rhs);
}

---

3) You mean an explicit specialization. Partial is still something else and also not possible for functions, only structs / classes. As such, since you can't specialize std::swap for template classes, you have to provide a free function in your namespace. Not a bad thing, if I may say so. Now, an explicit specialization is also possible, but generally you do not want to specialize a function template:

namespace std
{  // only allowed to extend namespace std with specializations

template<> // specialization
void swap<Bar>(Bar& lhs, Bar& rhs){
  // ...
}

}

---

4) No, as 1) is distinct from 2) and 3). Also, having both 2) and 3) will lead to always having 2) picked, because it fits better.

---

To answer the EDIT, where the classes may be template classes, you don't need specialization at all. consider a class like this:

template <class T>
struct vec3
{
    T x,y,z;
};

you may define classes such as:

vec3<float> a;
vec3<double> b;
vec3<int> c;

if you want to be able to create one function to implement all 3 swaps (not that this example class warrants it) you do just like Xeo said in (2)... without specialization but just make a regular template function:

template <class T>
void swap(vec3<T> &a, vec3<T> &b)
{
    using std::swap;
    swap(a.x,b.x);
    swap(a.y,b.y);
    swap(a.z,b.z);
}

The swap template function should be located in the same namespace as the class you're trying to swap. the following method will find and use that swap even though you're not referencing that namespace using ADL:

using std::swap;
swap(a,b);

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