C ++中的名称空间有效性

Question

I have a simple question for namespace in C++. There are errors when I compiling the following little piece of code. I don't understand why. Thanks for any help in advance!

#include <iostream>
using namespace std;

int x=100;

namespace first
{
     int x=1;
}
namespace second
{
     int x=2;
}

int main(){

     {
        using namespace first;
        cout<<x<<endl;
     }

     {
        using namespace second;
        cout<<x<<endl;
     }

     cout<<x<<endl;
}

If I comment out the x declared in the global scope and the last statement. It works fine. But in my mind, the first x is declared in the std namespace and using namespace first and second in the main will be invalid after the the code block they are declared(so the namespace will be std again). So the above code should work. Where am I wrong?

Answer 1

But in my mind, the first x is declared in the std namespace

Your mind is wrong. The using namespace std makes names from std available in the global namespace, it doesn't mean names declared in the global namespace are in std .

x is declared in the global namespace.

so the namespace will be std again

No, nothing in your file is in namespace std , only the contents of <iostream> are in namespace std .

The error is that when you try to use x there are two different variables in scope, ::x and first::x that you could be referring to, so it is ambiguous. You can disambiguate with a using declaration instead of a using directive:

 {
    using first::x;
    cout<<x<<endl;
 }

This says that in that scope x refers to first::x

Answer 2

in this case x variable is ambiguous. The compiler can't find which x you are going to use. You can write like this.

#include <iostream>
using namespace std;

int x=100;

namespace first
{
     int x=1;
}
namespace second
{
     int x=2;
}

int main(){

     {
        cout<<first::x<<endl;
     }

     {
        cout<<second::x<<endl;
     }

     cout<<x<<endl;
}

now your code will compile.

Answer 3

The first x is not in the std namespace. It would only be in the std namespace if defined like this:

namespace std
{
    int x;
}

Because it's not in that namespace, it's matched by references to x later in the program, and such references become ambiguous when another namespace declaring x is also being used (as in using namespace first / second).