Extending the namespace std

Adding declarations to std

It is undefined behavior to add declarations or definitions to namespace std or to any namespace nested within std, with a few exceptions noted below.

#include <utility>
 
namespace std
{
    // a function definition added to namespace std: undefined behavior
    pair<int, int> operator+(pair<int, int> a, pair<int, int> b)
    {
        return {a.first + b.first, a.second + b.second};
    }
}

Adding template specializations

Class templates

It is allowed to add template specializations for any standard library class template to the namespace std only if the declaration depends on at least one program-defined type and the specialization satisfies all requirements for the original template, except where such specializations are prohibited.

// Get the declaration of the primary std::hash template.
// We are not permitted to declare it ourselves.
// <typeindex> is guaranteed to provide such a declaration, 
// and is much cheaper to include than <functional>.
 
#include <typeindex> 
 
// Specialize std::hash so that MyType can be used as a key in 
// std::unordered_set and std::unordered_map.  Opening namespace
// std can accidentally introduce undefined behavior, and is not
// necessary for specializing class templates.
template<>
struct std::hash<MyType>
{
    std::size_t operator()(const MyType& t) const { return t.hash(); }
};

• Specializing the template std::complex for any type other than float, double, and long double is unspecified.
• Specializations of std::numeric_limits must define all members declared static const (until C++11)static constexpr (since C++11) in the primary template, in such a way that they are usable as integral constant expressions.

It is undefined behavior to declare a full or partial specialization of any member class template of a standard library class or class template.

Function templates and member functions of templates

It is undefined behavior to declare a full specialization of any member function of a standard library class template:

It is undefined behavior to declare a full specialization of any member function template of a standard library class or class template:

Variable templates

Explicit instantiation of templates

It is allowed to explicitly instantiate a class (since C++20)template defined in the standard library only if the declaration depends on the name of at least one program-defined type and the instantiation meets the standard library requirements for the original template.

Program-defined types

Program-defined specializations are explicit template specializations or partial specializations that are not part of the C++ standard library and not defined by the implementation.

Program-defined types are non-closure class types or enumeration types that are not part of the C++ standard library and not defined by the implementation, or closure type of non-implementation-provided lambda expressions (since C++11), or instantiation of program-defined specializations.

Other restrictions

The namespace std may not be declared as an inline namespace.

#include <cmath>
#include <memory>
 
int main()
{
    auto fptr0 = &static_cast<float(&)(float, float)>(std::betaf); // by unary operator&
    auto fptr1 = std::addressof(static_cast<float(&)(float, float)>(std::betaf)); // by std::addressof
    auto fptr2 = static_cast<float(&)(float)>(std::riemann_zetaf); // by function-to-pointer implicit conversion
    auto& fref = static_cast<float(&)(float)>(std::riemann_zetaf); // forming a reference
}

Defect reports

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.