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Dependent names

From cppreference.com
< cpp‎ | language

Inside the definition of a template (both class template and function template), the meaning of some constructs may differ from one instantiation to another. In particular, types and expressions may depend on types of type template parameters and values of non-type template parameters.

template<typename T>
struct X : B<T> // "B<T>" is dependent on T
{
    typename T::A* pa; // "T::A" is dependent on T
                       // (see below for the meaning of this use of "typename")
    void f(B<T>* pb) {
        static int i = B<T>::i; // "B<T>::i" is dependent on T
        pb->j++; // "pb->j" is dependent on T
    }
};

Name lookup and binding are different for dependent names and non-dependent names

Contents

[edit] Binding rules

Non-dependent names are looked up and bound at the point of template definition. This binding holds even if at the point of template instantiation there is a better match:

#include <iostream>
void g(double) { std::cout << "g(double)\n"; }
 
template<class T>
struct S {
    void f() const {
        g(1); // "g" is a non-dependent name, bound now
    }
};
 
void g(int) { std::cout << "g(int)\n"; }
 
int main()
{
    g(1); // calls g(int)
 
    S<int> s;
    s.f(); // calls g(double)
}


If the meaning of a non-dependent name changes between the definition context and the point of instantiation of a specialization of the template, the program is ill-formed, no diagnostic required. This is possible in the following situations:

  • a type used in a non-dependent name is incomplete at the point of definition but complete at the point of instantiation
  • (C++17) an instantiation uses a default argument or default template argument that had not been defined at the point of definition
  • (C++17) a constant expression at the point of instantiation uses the value of a const object of integral or unscoped enum type, the value of a constexpr object, the value of a reference, or the definition of a constexpr function, and that object/reference/function was not defined at the point of definition
  • (C++17) the template uses a non-dependent class template specialization or variable template specialization at the point of instantiation, and this template it uses is either instantiated from a partial specialization that was not defined at the point of definition or names an explicit specialization that was not declared at the point of definition

Binding of dependent names is postponed until lookup takes place.

[edit] Lookup rules

As discussed in lookup, the lookup of a dependent name used in a template is postponed until the template arguments are known, at which time

  • non-ADL lookup examines function declarations with external linkage that are visible from the template definition context
  • ADL examines function declarations with external linkage that are visible from both the template definition context and the template instantiation context

(in other words, adding a new function declaration after template definition does not make it visible, except via ADL).

The purpose of this is rule is to help guard against violations of the ODR for template instantiations:

// an external libary
namespace E {
  template<typename T>
  void writeObject(const T& t) {
    std::cout << "Value = " << t << '\n';
  }
}
 
// translation unit 1:
// Programmer 1 wants to allow E::writeObject to work with vector<int>
namespace P1 {
  std::ostream& operator<<(std::ostream& os, const std::vector<int>& v) {
      for(int n: v) os << n << ' '; return os;
  }
  void doSomething() {
    std::vector<int> v;
    E::writeObject(v); // error: will not find P1::operator<<
  }
}
 
// translation unit 2:
// Programmer 2 wants to allow E::writeObject to work with vector<int>
namespace P2 {
  std::ostream& operator<<(std::ostream& os, const std::vector<int>& v) {
      for(int n: v) os << n <<':'; return os << "[]";
  }
  void doSomethingElse() {
    std::vector<int> v;
    E::writeObject(v); // error: will not find P2::operator<<
  }
}

In the above example, if non-ADL lookup for operator<< were allowed from the instantiation context, the instantiation of E::writeObject<vector<int>> would have two different definitions: one using P1::operator<< and one using P2::operator<<. Such ODR violation may not be detected by the linker, leading to one or the other being used in both instances.

To make ADL examine a user-defined namespace, either std::vector should be replaced by a user-defined class or its element type should be a user-defined class:

namespace P1 {
  // if C is a class defined in the P1 namespace
  std::ostream& operator<<(std::ostream& os, const std::vector<C>& v) {
      for(C n: v) os << n; return os;
  }
  void doSomething() {
    std::vector<C> v;
    E::writeObject(v); // OK: instantiates writeObject(std::vector<P1::C>)
                       //     which finds P1::operator<< via ADL
  }
}

Note: this rule makes it impractical to overload operators for standard library types

#include <iostream>
#include <vector>
#include <iterator>
#include <utility>
 
// Bad idea: operator in global namespace, but its arguments are in std::
std::ostream& operator<<(std::ostream& os, std::pair<int, double> p)
{
    return os << p.first << ',' << p.second;
}
 
int main()
{
    typedef std::pair<int, double> elem_t;
    std::vector<elem_t> v(10);
    std::cout << v[0] << '\n'; // OK, ordinary lookup finds ::operator<<
    std::copy(v.begin(), v.end(),
              std::ostream_iterator<elem_t>(std::cout, " ")); // Error: both ordinary 
    // lookup from the point of definition of std::ostream_iterator and ADL will 
    // only consider the std namespace, and will find many overloads of
    // std::operator<<, so the lookup will be done. Overload resolution will then
    // fail to find operator<< for elem_t in the set found by the lookup.
}


Note: limited lookup (but not binding) of dependent names also takes place at template definition time, as needed to distinguish them from non-dependent names and also to determine whether they are members of the current instantiation or members of unknown specialization. The information obtained by this lookup can be used to detect errors, see below.

[edit] Dependent types

The following types are dependent types:

  • template parameter
  • a member of an unknown specialization (see below)
  • a nested class/enum that is a dependent member of unknown specialization (see below)
  • a cv-qualified version of a dependent type
  • a compound type constructed from a dependent type
  • an array type whose element type is dependent or whose bound (if any) is value-dependent
  • a template-id where either the template name is a template parameter, or any of template arguments is type- or value-dependent
  • the result of decltype applied to a type-dependent expression

Note: a typedef member of a current instantiation is only dependent when the type it refers to is.

[edit] Type-dependent expressions

The following expressions are type-dependent

  • an expression whose any subexpression is a type-dependent expression
  • this, if the class is a dependent type.
  • an id-expression that
  • contains an identifier for which name lookup finds at least one dependent declaration
  • contains a dependent template-id
  • contains the special identifier __func__ (if some enclosing function is a template, a non-template member of a class template, or a generic lambda (since C++14))
  • contains the name of conversion function to a dependent type
  • contains a nested name specifier or qualified-id that is a member of unknown specialization
  • names a dependent member of the current instantiation which is a static data member of type "array of unknown bound"
  • any cast expression to a dependent type
  • new-expression that creates an object of a dependent type
  • member access expression that refers to a member of the current instantiation whose type is dependent
  • member access expression that refers to a member of unknown specialization
(since C++17)

Note: literals, pseudo-destructor calls, sizeof, alignof, typeid, delete-expressions, throw-expressions, and noexcept-expressions are never type-dependent because the types of these expressions cannot be.

[edit] Value-dependent expressions

  • is a name declared with a dependent type
  • is a name of a non-type template parameter
  • names a member of unknown specialization
  • names a static data member that is a dependent member of the current instantiation and is not initialized.
  • names a static member function that is a depndent member of the current instantiation
(since C++14)
  • is a constant with a literal type, initialized from a value-dependent expression
  • sizeof, alignof, typeid, noexcept-expressions where the argument is a type-dependent expression or a dependent type-id
  • any cast expression to a dependent type or from a value-dependent expression
  • address-of expression where the argument is qualified-id that names a dependent member of the current instantiation
(since C++14)
(since C++17)

[edit] Dependent names

[edit] Current instantiation

Within a class template definition (including its member functions and nested classes) some names may be deduced to refer to the current instantiation. This allows certain errors to be detected at the point of definition, rather than instantiation, and removes the requirement on the typename and template disambiguators for dependent names, see below.

Only the following names can refer to the current instantiation:

  • in a class template definition:
  • a nested class, a member of class template, a member of a nested class, the injected-class-name of the template, the injected-class-name of a nested class
  • in a primary class template definition or in the definition of its member:
  • name of the class template followed by template argument list (or an equivalent alias template specialization) for the primary template where each argument is equivalent to its corresponding parameter. Note that if an expression is used as a non-type template argument (e.g. N+0 where N is the parameter), it does not name the current instantiation even if its value matches.
  • in the definition of a nested class or class template:
  • name of the nested class used as a member of the current instantiation
  • in the definition of a partial specialization or of a member of a partial specialization
  • the name of the class template followed by template argument list for the partial specialization, where each argument is equivalent to its corresponding parameter
template <class T> class A {
    A* p1;    // A is the current instantiation
    A<T>* p2; // A<T> is the current instantiation
    ::A<T>* p4; // ::A<T> is the current instantiation
    A<T*> p3; // A<T*> is not the current instantiation
    class B {
        B* p1; // B is the current instantiation
        A<T>::B* p2; // A<T>::B is the current instantiation
        typename A<T*>::B* p3; // A<T*>::B is not the current instantiation
    };
};
template <class T> class A<T*> {
    A<T*>* p1;  // A<T*> is the current instantiation
    A<T>* p2;   // A<T> is not the current instantiation
};
template <int I> struct B {
    static const int my_I = I;
    static const int my_I2 = I+0;
    static const int my_I3 = my_I;
    B<my_I>* b3;  // Bm<y_I> is the current instantiation
    B<my_I2>* b4; // B<my_I2> is not the current instantiation
    B<my_I3>* b5; // B<my_I3> is the current instantiation
};

Note that a base class can be the current instantiation if a nested class derives from its enclosing class template. Base classes that are dependent types but aren't the current instantiation are dependent base classes

template<class T> struct A {
    typedef int M;
    struct B {
        typedef void M;
        struct C;
    };
};
template<class T>
struct A<T>::B::C : A<T> {
    M m; // OK, A<T>::M
};

A name is classified as a member of the current instantiation if it is

  • an unqualified name that is found by unqualified lookup in the current instantiation or in its non-dependenent base.
  • qualified name, if the qualifier (the name to the left of ::) names the current instantiation and lookup finds the name in the current instantiation or in its non-dependent base
  • a name used in a class member access expression (y in x.y or xp->y), where the object expression (x or *xp) is the current instantiation and lookup finds the name in the current instantiation or in its non-dependent base
template <class T>
class A {
   static const int i = 5;
   int n1[i];       // i refers to a member of the current instantiation
   int n2[A::i];    // A::i refers to a member of the current instantiation
   int n3[A<T>::i]; // A<T>::i refers to a member of the current instantiation
   int f();
};
template <class T>
int A<T>::f() {
   return i; // i refers to a member of the current instantiation
}

Members of the current instantiation may be both dependent and non-dependent.

If the lookup of a member of current instantiation gives a different result between the point of instantiation and the point of definition, the lookup is ambiguous. Note however that when a member name is used, it is not automatically converted to a class member access expression, only explicit member access expressions indicate members of current instantiation:

struct A { int m; };
struct B { int m; };
 
template<typename T>
struct C : A, T {
  int f() { return this->m; }// finds A::m in the template definition context
  int g() { return m; }      // finds A::m in the template definition context
};
 
template int C<B>::f(); // error: finds both A::m and B::m
 
template int C<B>::g(); // OK: transformation to class member access syntax
                        // does not occur in the template definition context

[edit] Unknown specializations

Within a template definition, certain names are deduced to belong to an unknown specialization, in particular,

  • a qualified name, if any name that appears to the left of :: is a dependent type that is not a member of the current instantiation
  • a qualified name, whose qualifier is the current instantiation, and the name is not found in the current instantiation or any of its non-dependent base classes, and there is a dependent base class
  • a name of a member in a class member access expression (the y in x.y or xp->y), if the type of the object expression (x or *xp) is a dependent type and is not the current instantiation
  • a name of a member in a class member access expression (the y in x.y or xp->y), if the type of the object expression (x or *xp) is the current instantiation, and the name is not found in the current instantiation or any of its non-dependent base classes, and there is a dependent base class
template<typename T> struct Base {};
 
template<typename T>
struct Derived : Base<T> {
    void f() {
        // Derived<T> refers to current instantiation
        // there is no 'unknown_type' in the current instantiation
        // but there is a dependent base (Base<T>)
        // Therefore, unknown_type is a member of unknown specialization
        typename Derived<T>::unknown_type z;
    }
};
 
template<> struct Base<int> { // this specialization provides it
    typedef int unknown_type; 
};


This classification allows the following errors to be detected at the point of template definition (rather than instantiation):

  • If any template definition has a qualified name in which the qualifier refers to the current instantiation and the name is neither a member of current instantiation nor a member of unknown specialization, the program is ill-formed (no diagnostic required) even if the template is never instantiated.
template<class T>
class A {
  typedef int type;
  void f() {
    A<T>::type i; // OK: 'type' is a member of the current instantiation
    typename A<T>::other j; // Error:
    // 'other' is not a member of the current instantiation
    // and it is not a member of an unknown specialization
    // because A<T> (which names the current instantiation),
    // has no dependent bases for 'other' to hide in.
}
};
  • If any template definition has a member acess expression where the object expression is the current instantiation, but the name is neither a member of current instantiation nor a member of unknown specialization, the program is ill-formed even if the template is never instantiated.

Members of unknown specialization are always dependent, and are looked up and bound at the point of instantiation as all dependent names (see above)

[edit] The typename disambiguator for dependent names

In a declaration or a definition of a template, including alias template, a name that is not a member of the current instantiation and is dependent on a template parameter is not considered to be a type unless the keyword typename is used or unless it was already established as a type name, e.g. with a typedef declaration or by being used to name a base class.

#include <iostream>
#include <vector>
 
int p = 1;
template <typename T>
void foo(const std::vector<T> &v)
{
 
    // std::vector<T>::const_iterator is a dependent name,
    typename std::vector<T>::const_iterator it = v.begin();
 
    // without 'typename', the following is parsed as multiplication 
    // of the type-dependent member variable 'const_iterator' 
    // and some variable 'p'. Since there is a global 'p' visible
    // at this point, this template definition compiles.
    std::vector<T>::const_iterator* p; 
 
    typedef typename std::vector<T>::const_iterator iter_t;
    iter_t * p2; // iter_t is a dependent name, but it's known to be a type name
}
 
template<typename T>
struct S {
    typedef int value_t; // member of current instantiation
    void f() {
        S<T>::value_t n{};  // S<T> is dependenent, but 'typename' not needed
        std::cout << n << '\n';
    }
};
 
int main()
{
    std::vector<int> v;
    foo(v); // template instantiation fails: there is no member variable
            // called 'const_iterator' in the type std::vector<int>
    S<int>().f();
}


The keyword typename may only be used in this way before qualified names (e.g. T::x), but the names need not be dependent.

Usual qualified name lookup is used for the identifier prefixed by typename. Unlike the case with elaborated type specifier, the lookup rules do not change despite the qualifier:

struct A { // A has a nested variable X and a nested type struct X
   struct X {};
   int X;
};
struct B {
    struct X { }; // B has a nested type struct X
};
template<class T> void f(T t) {
    typename T::X x;
}
void foo() {
    A a;
    B b;
    f(b); // OK: instantiates f<B>, T::X refers to B::X
    f(a); // error: cannot instantiate f<A>:
          // because qualified name lookup for A::X finds the data member
}

The keyword typename must only be used in template declarations and definitions and only in contexts in which dependent names can be used. This excludes explicit specialization declarations and explicit instantiation declarations.

(until C++11)

The keyword typename can be used even outside of templates.

#include <vector>
 
int main() {
    typedef typename std::vector<T>::const_iterator iter_t; // OK in C++11
    typename std::vector<int> v; // also OK in C++11
 
}
(since C++11)

[edit] The template disambiguator for dependent names

Similarly, in a template definition, a dependent name that is not a member of the current instantiation is not considered to be a template name unless the disambiguation keyword template is used or unless it was already established as a template name:

template<typename T>
struct S {
    template<typename U> void foo(){}
};
 
template<typename T>
void bar()
{
    S<T> s;
    s.foo<T>(); // error: < parsed as less than operator
    s.template foo<T>(); // OK
}


The keyword template may only be used in this way after operators :: (scope resolution), -> (member access through pointer), and . (member access), the following are all valid examples:

  • T::template foo<X>();
  • s.template foo<X>();
  • this->template foo<X>();
  • typename T::template iterator<int>::value_type v;

As is the case with typename, the template prefix is allowed even if the name is not dependent or the use does not appear in the scope of a template (since C++11).

Even if the name to the left of :: refers to a namespace, the template disambiguator is allowed:

template<typename> struct s {};
::template s<void> q; // allowed, but unnecessary
(since C++17)

Due to the special rules for unqualified name lookup for template names in member access expressions, when a non-dependent template name appears in a member access expression (after -> or after .), the disambiguator is unnecessary if there is a class (since C++11) template with the same name found by ordinary lookup in the context of the expression. However, if the template found by lookup in the context of the expression differs from the one found in the context of the class, the program is ill-formed (until C++11)