Why is not std::swap in global namespace? - c++

Question

Item 25 in Effective c++ third edition, Scott Meyers suggests to implement swap in the same namespace as the class, and then when swapping to employ the using std::swap, and there the author says :
For example, if you were to write the call to swap this way:
std::swap(obj1,obj2); // the wrong way to call swap
you'd force the compiler to consider only the swap in std, thus
eliminating the possibility of getting a more appropriate T-specific
version defined elsewhere. Alas, some misguided programmers do qualify
calls to swap in this way, and that is why it's important to totally
specialize std::swap for your classes.
The author recommends to always swap objects this way :
#include <iostream>
#include <utility>
#define CUSTOM_SWAP
namespace aaa{
struct A
{
};
#ifdef CUSTOM_SWAP
void swap( A&, A& )
{
std::cout<<"not std::swap"<<std::endl;
}
#endif
}
int main()
{
using std::swap; // add std::swap to a list of possible resolutions
aaa::A a1;
aaa::A a2;
swap(a1,a2);
}
Why isn't std::swap in global namespace? That way, it would be simpler to add custom swap functions.

Answer 1

Probably because the standard says so, 17.6.1.1/2:
All library entities except macros, operator new and operator delete are defined within the namespace std or namespaces nested within namespace std.
And you would still need to put using ::swap sometimes, so it would introduce even more special cases. Here I use func instead of swap - http://ideone.com/WAWBfZ :
#include <iostream>
using namespace std;
template <class T>
auto func(T) -> void
{
cout << "::f" << endl;
}
namespace my_ns {
struct my_struct {};
auto func(my_struct) -> void
{
cout << "my_ns::func" << endl;
}
auto another_func() -> void
{
// won't compile without `using ::func;`
func(123);
}
}
auto main() -> int {}
fails with
prog.cpp: In function ‘void my_ns::another_func()’:
prog.cpp:21:17: error: could not convert ‘123’ from ‘int’ to ‘my_ns::my_struct’
func(123);