Why is "using namespace X;" not allowed at class/struct level?

Question

class C {
  using namespace std;  // error
};
namespace N {
  using namespace std; // ok
}
int main () {
  using namespace std; // ok
}

I want to know the motivation behind it.

Answer 1

I don't know exactly, but my guess is that allowing this at class scope could cause confusion:

namespace Hello
{
    typedef int World;
}

class Blah
{
    using namespace Hello;
public:
    World DoSomething();
}

//Should this be just World or Hello::World ?
World Blah::DoSomething()
{
    //Is the using namespace valid in here?
}

Since there is no obvious way of doing this, the standard just says you can't.

Now, the reason this is less confusing when we're talking namespace scopes:

namespace Hello
{
    typedef int World;
}

namespace Other
{
    using namespace Hello;
    World DoSomething();
}

//We are outside of any namespace, so we have to fully qualify everything. Therefore either of these are correct:

//Hello was imported into Other, so everything that was in Hello is also in Other. Therefore this is okay:
Other::World Other::DoSomething()
{
    //We're outside of a namespace; obviously the using namespace doesn't apply here.
    //EDIT: Apparently I was wrong about that... see comments. 
}

//The original type was Hello::World, so this is okay too.
Hello::World Other::DoSomething()
{
    //Ditto
}

namespace Other
{
    //namespace Hello has been imported into Other, and we are inside Other, so therefore we never need to qualify anything from Hello.
    //Therefore this is unambiguiously right
    World DoSomething()
    {
        //We're inside the namespace, obviously the using namespace does apply here.
    }
}

Answer 2

Because the C++ standard explicitly forbids it. From C++03 §7.3.4 [namespace.udir]:

using-directive:
    using namespace ::opt nested-name-specifieropt namespace-name ;

A using-directive shall not appear in class scope, but may appear in namespace scope or in block scope. [Note: when looking up a namespace-name in a using-directive, only namespace names are considered, see 3.4.6. ]

Why does the C++ standard forbid it? I don't know, ask a member of the ISO committee that approved the language standard.

Answer 3

I believe that the rationale is that it would probably be confusing. Currently, while processing a class level identifier, lookup will first search in the class scope and then in the enclosing namespace. Allowing the using namespace at class level would have quite some side effects on how the lookup is now performed. In particular, it would have to be performed sometime between checking that particular class scope and checking the enclosing namespace. That is: 1) merge the class level and used namespace level lookups, 2) lookup the used namespace after the class scope but before any other class scope, 3) lookup the used namespace right before the enclosing namespace. 4) lookup merged with the enclosing namespace.

1. This would make a big difference, where an identifier at class level would shadow any identifier in the enclosing namespace, but it would not shadow a used namespace. The effect would be strange, in that access to the used namespace from a class in a different namespace and from the same namespace would differ:

.

namespace A {
   void foo() {}
   struct B {
      struct foo {};
      void f() {
         foo();      // value initialize a A::B::foo object (current behavior)
      }
   };
}
struct C {
   using namespace A;
   struct foo {};
   void f() {
      foo();         // call A::foo
   }
};

1. Lookup right after this class scope. This would have the strange effect of shadowing base classes' members. The current lookup does not mix class and namespace level lookups, and when performing class lookup it will go all the way to the base classes before considering the enclosing namespace. The behavior would be surprising in that it would not consider the namespace in a similar level to the enclosing namespace. Again, the used namespace would be prioritized over the enclosing namespace.

.

namespace A {
   void foo() {}
}
void bar() {}
struct base {
   void foo();
   void bar();
};
struct test : base {
   using namespace A;
   void f() {
      foo();           // A::foo()
      bar();           // base::bar()
   }
};

1. Lookup right before the enclosing namespace. The problem with this approach is again that it would be surprising to many. Consider that the namespace is defined in a different translation unit, so that the following code cannot be seen all at once:

.

namespace A {
   void foo( int ) { std::cout << "int"; }
}
void foo( double ) { std::cout << "double"; }
struct test {
   using namespace A;
   void f() {
      foo( 5.0 );          // would print "int" if A is checked *before* the
                           // enclosing namespace
   }
};

1. Merge with the enclosing namespace. This would have the exact same effect that applying the using declaration at the namespace level. It would not add any new value to that, but will on the other hand complicate lookup for compiler implementors. Namespace identifier lookup is now independent from where in the code the lookup is triggered. When inside a class, if lookup does not find the identifier at class scope it will fall back to namespace lookup, but that is exactly the same namespace lookup that is used in a function definition, there is no need to maintain new state. When the using declaration is found at namespace level, the contents of the used namespace are brought into that namespace for all lookups involving the namespace. If using namespace was allowed at class level, there would be different outcomes for namespace lookup of the exact same namespace depending on where the lookup was triggered from, and that would make the implementation of the lookup much more complex for no additional value.

Anyway, my recommendation is not to employ the using namespace declaration at all. It makes code simpler to reason with without having to keep all namespaces' contents in mind.

Answer 4

This is probably disallowed because of openness vs closedness.

• Classes and structs in C++ are always closed entities. They are defined in exactly one place (although you can split declaration and implementation).
• namespaces can be opened, re-opened and extended arbitrarily often.

Importing namespaces into classes would lead to funny cases like this:

namespace Foo {}

struct Bar { using namespace Foo; };

namespace Foo {
using Baz = int; // I've just extended `Bar` with a type alias!
void baz(); // I've just extended `Bar` with what looks like a static function!
// etc.
}

Answer 5

I think it's a defect of the language. You may use workaround below. Keeping in mind this workaround, it is easy to suggest rules of names conflicts resolution for the case when the language will be changed.

namespace Hello
{
    typedef int World;
}
// surround the class (where we want to use namespace Hello)
// by auxiliary namespace (but don't use anonymous namespaces in h-files)
namespace Blah_namesp {
using namespace Hello;

class Blah
{
public:
    World DoSomething1();
    World DoSomething2();
    World DoSomething3();
};

World Blah::DoSomething1()
{
}

} // namespace Blah_namesp

// "extract" class from auxiliary namespace
using Blah_namesp::Blah;

Hello::World Blah::DoSomething2()
{
}
auto Blah::DoSomething3() -> World
{
}

Answer 6

You can't use using namespace inside of a class, but what you can do is simply use #define and then #undef inside of the structure. It will act the exact same way as namespace a = b;

struct foo
{
#define new_namespace old_namespace
     
    void foo2()
    {
        new_namespace::do_something();
    }

#undef new_namespace 
};