当外部名称空间的成员具有相同的名称时，访问未命名名称空间的成员

Question

Here is the test code

extern "C" {int printf(const char *, ...);}
namespace PS
{
   int x = 10; // A
   // some more code

   namespace {    
      int x = 20; // B
   }
   // more code
}

int main()
{
   printf("%d", PS::x); // prints 10
}

Is there any way to access inner(unnamed) namespace's x inside main ?

I dont want to change code inside PS . Apologies if the code looks highly impractical.

PS: I tend to use the name x quite often.

Answer 1

No. The only way to specify a namespace is by name, and the inner namespace has no name.

Assuming you can't rename either variable, you could reopen the inner namespace and add a differently-named accessor function or reference:

namespace PS {
    namespace {
        int & inner_x = x;
    }
}

printf("%d", PS::inner_x);

Answer 2

One way is to add this code:

namespace PS
{
   namespace
   {
      namespace access
      {
          int &xref = x;
      }
   }
}

and then you can access what you want:

std::cout << PS::access::xref << std::endl; //prints 20!

Demo : http://ideone.com/peqEs