namespace MyNamespace
{
static void foo1()
{
}
}
using namespace MyNamespace;
class MyClass
{
void foo2()
{
::foo1();
}
};
The scope resolution operation :: means using method in the global namespace. Here we can use ::foo1(). This means method foo1() is in the global namespace, am I right?
My question is, does using namespace ANAMESPACE_NAME mean we import all elements form the namespace ANAMESPACE_NAME into the global namespace?
No. "using namespace ANAMESPACE_NAME" mean we import all elements into current scope.
You can write something like this:
namespace A {
int i = 10, j = 20;
}
int f()
{
using namespace A; // injects names from A into the global scope.
return i * j; // uses i and j from namespace A.
}
int k = i * j; // Error: undefined variant i and j.
Section 3.4.3.4 of the C++2003 standard has an answer:
A name prefixed by the unary scope operator :: (5.1) is looked up in global scope, in the translation unit where it is used. The name shall be declared in global namespace scope or shall be a name whose declaration is visible in global scope because of a using-directive (3.4.3.2).
This paragraph is almost identical in the C++11 FDIS, so this probably also holds in C++11.
Here we can use ::foo1(). This means method foo1() is in the global namespace, am I right?
Yes thats correct. It means call the method named foo1() defined in global namespace.
This is called as Qualified Namespace Lookup.
do "using namespace ANAMESPACE_NAME" mean we import all elements in the ANAMESPACE_NAME namespace into global namespace?
Yes, it imports all elements from the ANAMESPACE_NAME namespace in to current namespace.
It is called as an using directive.
If you want to import just specific element in current type use, using declaration.
format is:
using ANAMESPACE_NAME::element_name;
