"네임 스페이스 X 사용"이유

"네임 스페이스 X 사용"이유 클래스 / 구조체 수준 내에서 허용되지 않습니까?

---

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

편집 : 동기를 알고 싶어요.

---

정확히 모르겠지만 클래스 범위에서 이것을 허용하면 혼란이 발생할 수 있다고 생각합니다.

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?
}

이를 수행하는 명백한 방법이 없기 때문에 표준은 할 수 없다고 말합니다.

이제 네임 스페이스 범위를 말할 때 이것이 덜 혼란스러운 이유는 다음과 같습니다.

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.
    }
}

---

C ++ 표준에서 명시 적으로 금지하기 때문입니다. C ++ 03 §7.3.4 [namespace.udir]에서 :

지시어 사용 :
    using namespace :: opt  nested-name-specifier opt  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.

---

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.

참고URL : https://stackoverflow.com/questions/6326805/why-using-namespace-x-is-not-allowed-inside-class-struct-level