10 Declarations [dcl.dcl]

10.3 Namespaces [basic.namespace]

A namespace is an optionally-named declarative region.
The name of a namespace can be used to access entities declared in that namespace; that is, the members of the namespace.
Unlike other declarative regions, the definition of a namespace can be split over several parts of one or more translation units.
The outermost declarative region of a translation unit is a namespace; see [basic.scope.namespace].

10.3.1 Namespace definition [namespace.def]

namespace-name:
	identifier
	namespace-alias
namespace-definition:
	named-namespace-definition
	unnamed-namespace-definition
	nested-namespace-definition
named-namespace-definition:
	inline namespace attribute-specifier-seq identifier { namespace-body }
unnamed-namespace-definition:
	inline namespace attribute-specifier-seq { namespace-body }
nested-namespace-definition:
	namespace enclosing-namespace-specifier :: identifier { namespace-body }
enclosing-namespace-specifier:
	identifier
	enclosing-namespace-specifier :: identifier
namespace-body:
	declaration-seq
Every namespace-definition shall appear in the global scope or in a namespace scope ([basic.scope.namespace]).
In a named-namespace-definition, the identifier is the name of the namespace.
If the identifier, when looked up, refers to a namespace-name (but not a namespace-alias) that was introduced in the namespace in which the named-namespace-definition appears or that was introduced in a member of the inline namespace set of that namespace, the namespace-definition extends the previously-declared namespace.
Otherwise, the identifier is introduced as a namespace-name into the declarative region in which the named-namespace-definition appears.
Because a namespace-definition contains declarations in its namespace-body and a namespace-definition is itself a declaration, it follows that namespace-definitions can be nested.
[Example
:
namespace Outer {
  int i;
  namespace Inner {
    void f() { i++; }           // Outer​::​i
    int i;
    void g() { i++; }           // Inner​::​i
  }
}
end example
]
The enclosing namespaces of a declaration are those namespaces in which the declaration lexically appears, except for a redeclaration of a namespace member outside its original namespace (e.g., a definition as specified in [namespace.memdef]).
Such a redeclaration has the same enclosing namespaces as the original declaration.
[Example
:
namespace Q {
  namespace V {
    void f();                   // enclosing namespaces are the global namespace, Q, and Q​::​V
    class C { void m(); };
  }
  void V::f() {                 // enclosing namespaces are the global namespace, Q, and Q​::​V
    extern void h();            // ... so this declares Q​::​V​::​h
  }
  void V::C::m() {              // enclosing namespaces are the global namespace, Q, and Q​::​V
  }
}
end example
]
If the optional initial inline keyword appears in a namespace-definition for a particular namespace, that namespace is declared to be an inline namespace.
The inline keyword may be used on a namespace-definition that extends a namespace only if it was previously used on the namespace-definition that initially declared the namespace-name for that namespace.
The optional attribute-specifier-seq in a named-namespace-definition appertains to the namespace being defined or extended.
Members of an inline namespace can be used in most respects as though they were members of the enclosing namespace.
Specifically, the inline namespace and its enclosing namespace are both added to the set of associated namespaces used in argument-dependent lookup whenever one of them is, and a using-directive that names the inline namespace is implicitly inserted into the enclosing namespace as for an unnamed namespace.
Furthermore, each member of the inline namespace can subsequently be partially specialized, explicitly instantiated, or explicitly specialized as though it were a member of the enclosing namespace.
Finally, looking up a name in the enclosing namespace via explicit qualification ([namespace.qual]) will include members of the inline namespace brought in by the using-directive even if there are declarations of that name in the enclosing namespace.
These properties are transitive: if a namespace N contains an inline namespace M, which in turn contains an inline namespace O, then the members of O can be used as though they were members of M or N.
The inline namespace set of N is the transitive closure of all inline namespaces in N.
The enclosing namespace set of O is the set of namespaces consisting of the innermost non-inline namespace enclosing an inline namespace O, together with any intervening inline namespaces.
A nested-namespace-definition with an enclosing-namespace-specifier E, identifier I and namespace-body B is equivalent to
namespace E { namespace I { B } }
[Example
:
namespace A::B::C {
  int i;
}
The above has the same effect as:
namespace A {
  namespace B {
    namespace C {
      int i;
    }
  }
}
end example
]

10.3.1.1 Unnamed namespaces [namespace.unnamed]

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.
[Example
:
namespace { int i; }            // unique​::​i
void f() { i++; }               // unique​::​i++

namespace A {
  namespace {
    int i;                      // A​::​unique​::​i
    int j;                      // A​::​unique​::​j
  }
  void g() { i++; }             // A​::​unique​::​i++
}

using namespace A;
void h() {
  i++;                          // error: unique​::​i or A​::​unique​::​i
  A::i++;                       // A​::​unique​::​i
  j++;                          // A​::​unique​::​j
}
end example
]

10.3.1.2 Namespace member definitions [namespace.memdef]

A declaration in a namespace N (excluding declarations in nested scopes) whose declarator-id is an unqualified-id ([dcl.meaning]), whose class-head-name or enum-head-name is an identifier, or whose elaborated-type-specifier is of the form class-key attribute-specifier-seq identifier ([dcl.type.elab]), or that is an opaque-enum-declaration, declares (or redeclares) its unqualified-id or identifier as a member of N.
[Note
:
An explicit instantiation or explicit specialization of a template does not introduce a name and thus may be declared using an unqualified-id in a member of the enclosing namespace set, if the primary template is declared in an inline namespace.
end note
]
[Example
:
namespace X {
  void f() { /* ... */ }        // OK: introduces X​::​f()

  namespace M {
    void g();                   // OK: introduces X​::​M​::​g()
  }
  using M::g;
  void g();                     // error: conflicts with X​::​M​::​g()
}
end example
]
Members of a named namespace can also be defined outside that namespace by explicit qualification ([namespace.qual]) of the name being defined, provided that the entity being defined was already declared in the namespace and the definition appears after the point of declaration in a namespace that encloses the declaration's namespace.
[Example
:
namespace Q {
  namespace V {
    void f();
  }
  void V::f() { /* ... */ }     // OK
  void V::g() { /* ... */ }     // error: g() is not yet a member of V
  namespace V {
    void g();
  }
}

namespace R {
  void Q::V::g() { /* ... */ }  // error: R doesn't enclose Q
}
end example
]
If a friend declaration in a non-local class first declares a class, function, class template or function template99 the friend is a member of the innermost enclosing namespace.
The friend declaration does not by itself make the name visible to unqualified lookup or qualified lookup.
[Note
:
The name of the friend will be visible in its namespace if a matching declaration is provided at namespace scope (either before or after the class definition granting friendship).
end note
]
If a friend function or function template is called, its name may be found by the name lookup that considers functions from namespaces and classes associated with the types of the function arguments ([basic.lookup.argdep]).
If the name in a friend declaration is neither qualified nor a template-id and the declaration is a function or an elaborated-type-specifier, the lookup to determine whether the entity has been previously declared shall not consider any scopes outside the innermost enclosing namespace.
[Note
:
The other forms of friend declarations cannot declare a new member of the innermost enclosing namespace and thus follow the usual lookup rules.
end note
]
[Example
:
// Assume f and g have not yet been declared.
void h(int);
template <class T> void f2(T);
namespace A {
  class X {
    friend void f(X);           // A​::​f(X) is a friend
    class Y {
      friend void g();          // A​::​g is a friend
      friend void h(int);       // A​::​h is a friend
                                // ​::​h not considered
      friend void f2<>(int);    // ​::​f2<>(int) is a friend
    };
  };

  // A​::​f, A​::​g and A​::​h are not visible here
  X x;
  void g() { f(x); }            // definition of A​::​g
  void f(X) { /* ... */ }       // definition of A​::​f
  void h(int) { /* ... */ }     // definition of A​::​h
  // A​::​f, A​::​g and A​::​h are visible here and known to be friends
}

using A::x;

void h() {
  A::f(x);
  A::X::f(x);                   // error: f is not a member of A​::​X
  A::X::Y::g();                 // error: g is not a member of A​::​X​::​Y
}
end example
]
this implies that the name of the class or function is unqualified.

10.3.2 Namespace alias [namespace.alias]

The identifier in a namespace-alias-definition is a synonym for the name of the namespace denoted by the qualified-namespace-specifier and becomes a namespace-alias.
[Note
:
When looking up a namespace-name in a namespace-alias-definition, only namespace names are considered, see [basic.lookup.udir].
end note
]
In a declarative region, a namespace-alias-definition can be used to redefine a namespace-alias declared in that declarative region to refer only to the namespace to which it already refers.
[Example
:
The following declarations are well-formed:
namespace Company_with_very_long_name { /* ... */ }
namespace CWVLN = Company_with_very_long_name;
namespace CWVLN = Company_with_very_long_name;  // OK: duplicate
namespace CWVLN = CWVLN;
end example
]

10.3.3 The using declaration [namespace.udecl]

using-declaration:
	using using-declarator-list ;
using-declarator-list:
	using-declarator ...
	using-declarator-list , using-declarator ...
using-declarator:
	typename nested-name-specifier unqualified-id
Each using-declarator in a using-declaration100 introduces a set of declarations into the declarative region in which the using-declaration appears.
The set of declarations introduced by the using-declarator is found by performing qualified name lookup ([basic.lookup.qual], [class.member.lookup]) for the name in the using-declarator, excluding functions that are hidden as described below.
If the using-declarator does not name a constructor, the unqualified-id is declared in the declarative region in which the using-declaration appears as a synonym for each declaration introduced by the using-declarator.
[Note
:
Only the specified name is so declared; specifying an enumeration name in a using-declaration does not declare its enumerators in the using-declaration's declarative region.
end note
]
If the using-declarator names a constructor, it declares that the class inherits the set of constructor declarations introduced by the using-declarator from the nominated base class.
Every using-declaration is a declaration and a member-declaration and can therefore be used in a class definition.
[Example
:
struct B {
  void f(char);
  void g(char);
  enum E { e };
  union { int x; };
};

struct D : B {
  using B::f;
  void f(int) { f('c'); }       // calls B​::​f(char)
  void g(int) { g('c'); }       // recursively calls D​::​g(int)
};
end example
]
In a using-declaration used as a member-declaration, each using-declarator's nested-name-specifier shall name a base class of the class being defined.
If a using-declarator names a constructor, its nested-name-specifier shall name a direct base class of the class being defined.
[Example
:
template <typename... bases>
struct X : bases... {
  using bases::g...;
};

X<B, D> x;                      // OK: B​::​g and D​::​g introduced
end example
]
[Example
:
class C {
  int g();
};

class D2 : public B {
  using B::f;                   // OK: B is a base of D2
  using B::e;                   // OK: e is an enumerator of base B
  using B::x;                   // OK: x is a union member of base B
  using C::g;                   // error: C isn't a base of D2
};
end example
]
[Note
:
Since destructors do not have names, a using-declaration cannot refer to a destructor for a base class.
Since specializations of member templates for conversion functions are not found by name lookup, they are not considered when a using-declaration specifies a conversion function ([temp.mem]).
end note
]
If a constructor or assignment operator brought from a base class into a derived class has the signature of a copy/move constructor or assignment operator for the derived class ([class.copy]), the using-declaration does not by itself suppress the implicit declaration of the derived class member; the member from the base class is hidden or overridden by the implicitly-declared copy/move constructor or assignment operator of the derived class, as described below.
A using-declaration shall not name a template-id.
[Example
:
struct A {
  template <class T> void f(T);
  template <class T> struct X { };
};
struct B : A {
  using A::f<double>;           // ill-formed
  using A::X<int>;              // ill-formed
};
end example
]
A using-declaration shall not name a namespace.
A using-declaration shall not name a scoped enumerator.
A using-declaration that names a class member shall be a member-declaration.
[Example
:
struct X {
  int i;
  static int s;
};

void f() {
  using X::i;                   // error: X​::​i is a class member and this is not a member declaration.
  using X::s;                   // error: X​::​s is a class member and this is not a member declaration.
}
end example
]
Members declared by a using-declaration can be referred to by explicit qualification just like other member names ([namespace.qual]).
[Example
:
void f();

namespace A {
  void g();
}

namespace X {
  using ::f;                    // global f
  using A::g;                   // A's g
}

void h()
{
  X::f();                       // calls ​::​f
  X::g();                       // calls A​::​g
}
end example
]
A using-declaration is a declaration and can therefore be used repeatedly where (and only where) multiple declarations are allowed.
[Example
:
namespace A {
  int i;
}

namespace A1 {
  using A::i, A::i;             // OK: double declaration
}

struct B {
  int i;
};

struct X : B {
  using B::i, B::i;             // error: double member declaration
};
end example
]
[Note
:
For a using-declaration whose nested-name-specifier names a namespace, members added to the namespace after the using-declaration are not in the set of introduced declarations, so they are not considered when a use of the name is made.
Thus, additional overloads added after the using-declaration are ignored, but default function arguments ([dcl.fct.default]), default template arguments ([temp.param]), and template specializations ([temp.class.spec], [temp.expl.spec]) are considered.
end note
]
[Example
:
namespace A {
  void f(int);
}

using A::f;         // f is a synonym for A​::​f; that is, for A​::​f(int).
namespace A {
  void f(char);
}

void foo() {
  f('a');           // calls f(int), even though f(char) exists.
}

void bar() {
  using A::f;       // f is a synonym for A​::​f; that is, for A​::​f(int) and A​::​f(char).
  f('a');           // calls f(char)
}
end example
]
[Note
:
Partial specializations of class templates are found by looking up the primary class template and then considering all partial specializations of that template.
If a using-declaration names a class template, partial specializations introduced after the using-declaration are effectively visible because the primary template is visible ([temp.class.spec]).
end note
]
Since a using-declaration is a declaration, the restrictions on declarations of the same name in the same declarative region also apply to using-declarations.
[Example
:
namespace A {
  int x;
}

namespace B {
  int i;
  struct g { };
  struct x { };
  void f(int);
  void f(double);
  void g(char);     // OK: hides struct g
}

void func() {
  int i;
  using B::i;       // error: i declared twice
  void f(char);
  using B::f;       // OK: each f is a function
  f(3.5);           // calls B​::​f(double)
  using B::g;
  g('a');           // calls B​::​g(char)
  struct g g1;      // g1 has class type B​::​g
  using B::x;
  using A::x;       // OK: hides struct B​::​x
  x = 99;           // assigns to A​::​x
  struct x x1;      // x1 has class type B​::​x
}
end example
]
If a function declaration in namespace scope or block scope has the same name and the same parameter-type-list as a function introduced by a using-declaration, and the declarations do not declare the same function, the program is ill-formed.
If a function template declaration in namespace scope has the same name, parameter-type-list, return type, and template parameter list as a function template introduced by a using-declaration, the program is ill-formed.
[Note
:
Two using-declarations may introduce functions with the same name and the same parameter-type-list.
If, for a call to an unqualified function name, function overload resolution selects the functions introduced by such using-declarations, the function call is ill-formed.
[Example
:
namespace B {
  void f(int);
  void f(double);
}
namespace C {
  void f(int);
  void f(double);
  void f(char);
}

void h() {
  using B::f;       // B​::​f(int) and B​::​f(double)
  using C::f;       // C​::​f(int), C​::​f(double), and C​::​f(char)
  f('h');           // calls C​::​f(char)
  f(1);             // error: ambiguous: B​::​f(int) or C​::​f(int)?
  void f(int);      // error: f(int) conflicts with C​::​f(int) and B​::​f(int)
}
end example
]
end note
]
When a using-declarator brings declarations from a base class into a derived class, member functions and member function templates in the derived class override and/or hide member functions and member function templates with the same name, parameter-type-list, cv-qualification, and ref-qualifier (if any) in a base class (rather than conflicting).
Such hidden or overridden declarations are excluded from the set of declarations introduced by the using-declarator.
[Example
:
struct B {
  virtual void f(int);
  virtual void f(char);
  void g(int);
  void h(int);
};

struct D : B {
  using B::f;
  void f(int);      // OK: D​::​f(int) overrides B​::​f(int);

  using B::g;
  void g(char);     // OK

  using B::h;
  void h(int);      // OK: D​::​h(int) hides B​::​h(int)
};

void k(D* p)
{
  p->f(1);          // calls D​::​f(int)
  p->f('a');        // calls B​::​f(char)
  p->g(1);          // calls B​::​g(int)
  p->g('a');        // calls D​::​g(char)
}

struct B1 {
  B1(int);
};

struct B2 {
  B2(int);
};

struct D1 : B1, B2 {
  using B1::B1;
  using B2::B2;
};
D1 d1(0);           // ill-formed: ambiguous

struct D2 : B1, B2 {
  using B1::B1;
  using B2::B2;
  D2(int);          // OK: D2​::​D2(int) hides B1​::​B1(int) and B2​::​B2(int)
};
D2 d2(0);           // calls D2​::​D2(int)
end example
]
For the purpose of forming a set of candidates during overload resolution, the functions that are introduced by a using-declaration into a derived class are treated as though they were members of the derived class.
In particular, the implicit this parameter shall be treated as if it were a pointer to the derived class rather than to the base class.
This has no effect on the type of the function, and in all other respects the function remains a member of the base class.
Likewise, constructors that are introduced by a using-declaration are treated as though they were constructors of the derived class when looking up the constructors of the derived class ([class.qual]) or forming a set of overload candidates ([over.match.ctor], [over.match.copy], [over.match.list]).
If such a constructor is selected to perform the initialization of an object of class type, all subobjects other than the base class from which the constructor originated are implicitly initialized ([class.inhctor.init]).
[Note
:
A member of a derived class is sometimes preferred to a member of a base class if they would otherwise be ambiguous ([over.match.best]).
end note
]
In a using-declarator that does not name a constructor, all members of the set of introduced declarations shall be accessible.
In a using-declarator that names a constructor, no access check is performed.
In particular, if a derived class uses a using-declarator to access a member of a base class, the member name shall be accessible.
If the name is that of an overloaded member function, then all functions named shall be accessible.
The base class members mentioned by a using-declarator shall be visible in the scope of at least one of the direct base classes of the class where the using-declarator is specified.
[Note
:
Because a using-declarator designates a base class member (and not a member subobject or a member function of a base class subobject), a using-declarator cannot be used to resolve inherited member ambiguities.
[Example
:
struct A { int x(); };
struct B : A { };
struct C : A {
  using A::x;
  int x(int);
};

struct D : B, C {
  using C::x;
  int x(double);
};
int f(D* d) {
  return d->x();    // error: overload resolution selects A​::​x, but A is an ambiguous base class
}
end example
]
end note
]
A synonym created by a using-declaration has the usual accessibility for a member-declaration.
A using-declarator that names a constructor does not create a synonym; instead, the additional constructors are accessible if they would be accessible when used to construct an object of the corresponding base class, and the accessibility of the using-declaration is ignored.
[Example
:
class A {
private:
    void f(char);
public:
    void f(int);
protected:
    void g();
};

class B : public A {
  using A::f;       // error: A​::​f(char) is inaccessible
public:
  using A::g;       // B​::​g is a public synonym for A​::​g
};
end example
]
If a using-declarator uses the keyword typename and specifies a dependent name ([temp.dep]), the name introduced by the using-declaration is treated as a typedef-name.
A using-declaration with more than one using-declarator is equivalent to a corresponding sequence of using-declarations with one using-declarator each.

10.3.4 Using directive [namespace.udir]

using-directive:
	attribute-specifier-seq using  namespace nested-name-specifier 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 [basic.lookup.udir].
end note
]
The optional attribute-specifier-seq appertains to the using-directive.
A using-directive specifies that the names in the nominated namespace can be used in the scope in which the using-directive appears after the using-directive.
During unqualified name lookup, the names appear as if they were declared in the nearest enclosing namespace which contains both the using-directive and the nominated namespace.
[Note
:
In this context, “contains” means “contains directly or indirectly”.
end note
]
A using-directive does not add any members to the declarative region in which it appears.
[Example
:
namespace A {
  int i;
  namespace B {
    namespace C {
      int i;
    }
    using namespace A::B::C;
    void f1() {
      i = 5;        // OK, C​::​i visible in B and hides A​::​i
    }
  }
  namespace D {
    using namespace B;
    using namespace C;
    void f2() {
      i = 5;        // ambiguous, B​::​C​::​i or A​::​i?
    }
  }
  void f3() {
    i = 5;          // uses A​::​i
  }
}
void f4() {
  i = 5;            // ill-formed; neither i is visible
}
end example
]
For unqualified lookup, 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.
[Note
:
For qualified lookup, see [namespace.qual].
end note
]
[Example
:
namespace M {
  int i;
}

namespace N {
  int i;
  using namespace M;
}

void f() {
  using namespace N;
  i = 7;            // error: both M​::​i and N​::​i are visible
}
For another example,
namespace A {
  int i;
}
namespace B {
  int i;
  int j;
  namespace C {
    namespace D {
      using namespace A;
      int j;
      int k;
      int a = i;    // B​::​i hides A​::​i
    }
    using namespace D;
    int k = 89;     // no problem yet
    int l = k;      // ambiguous: C​::​k or D​::​k
    int m = i;      // B​::​i hides A​::​i
    int n = j;      // D​::​j hides B​::​j
  }
}
end example
]
If a namespace is extended after a using-directive for that namespace is given, the additional members of the extended namespace and the members of namespaces nominated by using-directives in the extending namespace-definition can be used after the extending namespace-definition.
If name lookup finds a declaration for a name in two different namespaces, and the declarations do not declare the same entity and do not declare functions, the use of the name is ill-formed.
[Note
:
In particular, the name of a variable, function or enumerator does not hide the name of a class or enumeration declared in a different namespace.
For example,
namespace A {
  class X { };
  extern "C"   int g();
  extern "C++" int h();
}
namespace B {
  void X(int);
  extern "C"   int g();
  extern "C++" int h(int);
}
using namespace A;
using namespace B;

void f() {
  X(1);             // error: name X found in two namespaces
  g();              // OK: name g refers to the same entity
  h();              // OK: overload resolution selects A​::​h
}
end note
]
During overload resolution, all functions from the transitive search are considered for argument matching.
The set of declarations found by the transitive search is unordered.
[Note
:
In particular, the order in which namespaces were considered and the relationships among the namespaces implied by the using-directives do not cause preference to be given to any of the declarations found by the search.
end note
]
An ambiguity exists if the best match finds two functions with the same signature, even if one is in a namespace reachable through using-directives in the namespace of the other.101
[Example
:
namespace D {
  int d1;
  void f(char);
}
using namespace D;

int d1;             // OK: no conflict with D​::​d1

namespace E {
  int e;
  void f(int);
}

namespace D {       // namespace extension
  int d2;
  using namespace E;
  void f(int);
}

void f() {
  d1++;             // error: ambiguous ​::​d1 or D​::​d1?
  ::d1++;           // OK
  D::d1++;          // OK
  d2++;             // OK: D​::​d2
  e++;              // OK: E​::​e
  f(1);             // error: ambiguous: D​::​f(int) or E​::​f(int)?
  f('a');           // OK: D​::​f(char)
}
end example
]
During name lookup in a class hierarchy, some ambiguities may be resolved by considering whether one member hides the other along some paths ([class.member.lookup]).
There is no such disambiguation when considering the set of names found as a result of following using-directives.