Scope of `using namespace` in Unnamed Namespace

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Why is the using directive using namespace std; in when enclosed in an anonymous namespace behaves as if it appeared at global scope?

#include <iostream>

namespace x
{
  using namespace std;
  void g()
  {
    cout << 1;
  }
}

int main()
{
  cout << 1; // compiles fine if `namespace x` is replaced with `namespace`
}
using-directives namespaces c++
1 Answers

An unnamed namespace is equivalent to essentially writing:

namespace __compiler_generated_unique {

}
using namespace __compiler_generated_unique;

So it's like having a using directive at global scope. And using directives are transitive.

For a normative reference, here it is from n4861 (The C++20 standard draft):

[namespace.unnamed]

1 An unnamed-namespace-definition behaves as if it were replaced by

inline namespace unique { /* empty body */ }
using namespace unique ;
namespace unique { namespace-body }

where inline appears if and only if it appears in the unnamed-namespace-definition and all occurrences of unique in a translation unit are replaced by the same identifier, and this identifier differs from all other identifiers in the translation unit. The optional attribute-specifier-seq in the unnamed-namespace-definition appertains to unique.

[namespace.udir]

4 For unqualified lookup ([basic.lookup.unqual]), the using-directive is transitive: if a scope contains a using-directive that nominates a second namespace that itself contains using-directives, the effect is as if the using-directives from the second namespace also appeared in the first.

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