Call of overloaded function is ambiguous although different namespace - c++

Question

I do understand why the following would be a problem if no namespaces were used. The call would be ambiguous indeed. I thought "using stD::swap;" would define which method to use.
Why does it work for "int" but not a "class"?
#include <memory>
namespace TEST {
class Dummy{};
void swap(Dummy a){};
void sw(int x){};
}
namespace stD {
void swap(TEST::Dummy a){};
void sw(int x){};
class aClass{
public:
void F()
{
using stD::swap;
TEST::Dummy x;
swap(x);
}
void I()
{
using stD::sw;
int b = 0;
sw(b);
}
};
}
This is the error message:
../src/Test.h: In member function ‘void stD::aClass::F()’:
../src/Test.h:26:9: error: call of overloaded ‘swap(TEST::Dummy&)’ is ambiguous
swap(x);
^
../src/Test.h:26:9: note: candidates are:
../src/Test.h:17:6: note: void stD::swap(TEST::Dummy)
void swap(TEST::Dummy a){};
^
../src/Test.h:10:6: note: void TEST::swap(TEST::Dummy)
void swap(Dummy a){};
^
I thank you very much in advance for an answer.

Answer 1

This line is using argument dependent lookup
TEST::Dummy x;
swap(x);
So it will find both void stD::swap(TEST::Dummy) as well as void TEST::swap(TEST::Dummy) because x carries the TEST:: namespace.
In the latter case int b = 0; the variable b is not in a namespace, so the only valid function to call would be stD::sw due to your using statement.