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.

Solution 1:

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. Here 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) noexcept {
    // ...
}

}

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.

Solution 2:

It seems that (2) (free standing swap in the same namespace where the user-defined class is declared) is the only allowed way to provide swap for a user-defined class, because adding declarations to namespace std is generally an undefined behavior. Extending the namespace std (cppreference.com):

It is undefined behavior to add declarations or definitions to namespace std or to any namespace nested within std, with a few exceptions noted below

And swap is not denoted as one of those exceptions. So adding your own swap overload to the std namespace is an undefined behavior.

It's also said that the standard library uses an unqualified call to the swap function in order to call user-defined swap for a user class if such user-defined swap is provided.

Swappable (cppreference.com):

Many standard library functions (for example, many algorithms) expect their arguments to satisfy Swappable, which means that any time the standard library performs a swap, it uses the equivalent of using std::swap; swap(t, u);.

swap (www.cplusplus.com):

Many components of the standard library (within std) call swap in an unqualified manner to allow custom overloads for non-fundamental types to be called instead of this generic version: Custom overloads of swap declared in the same namespace as the type for which they are provided get selected through argument-dependent lookup over this generic version.

But note that directly using the std::swap function for a user-defined class calls the generic version of std::swap instead of the user-defined swap:

my::object a, b;
std::swap(a, b); // calls std::swap, not my::swap

So it is recommended to call the swap function in user code in the same way as it is done in the standard library:

my::object a, b;
using std::swap;
swap(a, b); // calls my::swap if it is defined, or std::swap if it is not.

Solution 3:

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);