what does `using std::swap` inside the body of a class method implementation mean?

Question

I was trying to learn and adopt the copy-swap idiom following this thorough explanation on this question: the Copy-Swap Idiom.

But I found some code I had never seen: using std::swap; // allow ADL in this example

class dumb_array { public:     // ...      void swap(dumb_array& pOther) // nothrow     {         using std::swap; // allow ADL    /* <===== THE LINE I DONT UNDERSTAND */          swap(mSize, pOther.mSize); // with the internal members swapped,         swap(mArray, pOther.mArray); // *this and pOther are effectively swapped     } }; 

1. what does using std::swap; mean inside the body of a function implementation ?
2. what does ADL mean ?

Answer 1

This mechanism is normally used in templated code, i.e. template <typename Value> class Foo.

Now the question is which swap to use. std::swap<Value> will work, but it might not be ideal. There's a good chance that there's a better overload of swap for type Value, but in which namespace would that be? It's almost certainly not in std:: (since that's illegal), but quite likely in the namespace of Value. Likely, but far from certain.

In that case, swap(myValue, anotherValue) will get you the "best" swap possible. Argument Dependent Lookup will find any swap in the namespace where Value came from. Otherwise the using directive kicks in, and std::swap<Value> will be instantiated and used.

In your code, mSize is likely an integral type, and mArray a pointer. Neither has an associated namespace, and std::swap is with 99.9% certainty optimal for them anyway. Therefore, the using std::swap; declaration seems useless here.

Answer 2

The using keyword has scoped effect.

This means that std::swap can be referred to as swap during the scope of the using keyword.