std::common_type (3) - Linux Manuals

std::common_type: std::common_type


std::common_type - std::common_type


Defined in header <type_traits>
template< class... T > (since C++11)
struct common_type;

Determines the common type among all types T..., that is the type all T... can be implicitly converted to. If such a type exists (as determined according to the rules below), the member type names that type. Otherwise, there is no member type.

* If sizeof...(T) is zero, there is no member type.
* If sizeof...(T) is one (i.e., T... contains only one type T0), the member type names the same type as std::common_type<T0, T0>::type if it exists; otherwise there is no member type.
* If sizeof...(T) is two (i.e., T... contains exactly two types T1 and T2),

      * If applying std::decay to at least one of T1 and T2 produces a different type, the member type names the same type as std::common_type<std::decay<T1>::type, std::decay<T2>::type>::type, if it exists; if not, there is no member type.
      * Otherwise, if there is a user specialization for std::common_type<T1, T2>, that specialization is used;
      * Otherwise, if std::decay<decltype(false ? std::declval<T1>() : std::declval<T2>())>::type is a valid type, the member type denotes that type;

      * Otherwise, if std::decay<decltype(false ? std::declval<const T1 &>() : std::declval<const T2 &>())>::type is a valid type, the member type denotes that type; (since C++20)

      * Otherwise, there is no member type.

* If sizeof...(T) is greater than two (i.e., T... consists of the types T1, T2, R...), then if std::common_type<T1, T2>::type exists, the member type denotes std::common_type<std::common_type<T1, T2>::type, R...>::type if such a type exists. In all other cases, there is no member type.

The types in the parameter pack T shall each be a complete type, (possibly cv-qualified) void, or an array of unknown bound. Otherwise, the behavior is undefined.
If an instantiation of a template above depends, directly or indirectly, on an incomplete type, and that instantiation could yield a different result if that type were hypothetically completed, the behavior is undefined.

Member types

Name Definition
type the common type for all T...

Helper types

template< class... T > (since C++14)
using common_type_t = typename common_type<T...>::type;


Users may specialize common_type for types T1 and T2 if

* At least one of T1 and T2 depends on a user-defined type, and
* std::decay is an identity transformation for both T1 and T2.

If such a specialization has a member named type, it must be a public and unambiguous member type that names a cv-unqualified non-reference type to which both T1 and T2 are explicitly convertible. Additionally, std::common_type<T1, T2>::type and std::common_type<T2, T1>::type must denote the same type.
A program that adds common_type specializations in violation of these rules has undefined behavior.
Note that the behavior of a program that adds a specialization to any other template from <type_traits> is undefined.
The following specializations are already provided by the standard library:

                                          specializes the std::common_type trait
std::common_type<std::chrono::duration> (class template specialization)
                                          specializes the std::common_type trait
std::common_type<std::chrono::time_point> (class template specialization)

Possible implementation

  // primary template (used for zero types)
  template <class ...T>
  struct common_type {};

  //////// one type
  template <class T>
  struct common_type<T> : common_type<T, T> {};

  //////// two types

  // default implementation for two types
  template<class T1, class T2>
  using cond_t = decltype(false ? std::declval<T1>() : std::declval<T2>());

  template<class T1, class T2, class=void>
  struct common_type_2_default {};

  template<class T1, class T2>
  struct common_type_2_default<T1, T2, std::void_t<cond_t<T1, T2>>> {
      using type = std::decay_t<cond_t<T1, T2>>;

  // dispatcher to decay the type before applying specializations
  template<class T1, class T2, class D1 = std::decay_t<T1>, class D2=std::decay_t<T2>>
  struct common_type_2_impl : common_type<D1, D2> {};

  template<class D1, class D2>
  struct common_type_2_impl<D1, D2, D1, D2> : common_type_2_default<D1, D2> {};

  template <class T1, class T2>
  struct common_type<T1, T2> : common_type_2_impl<T1, T2> { };

  //////// 3+ types

  template<class AlwaysVoid, class T1, class T2, class...R>
  struct common_type_multi_impl { };

  template< class T1, class T2, class...R>
  struct common_type_multi_impl<std::void_t<common_type_t<T1, T2>>, T1, T2, R...>
      : common_type<common_type_t<T1, T2>, R...> { };

  template <class T1, class T2, class... R>
  struct common_type<T1, T2, R...>
      : common_type_multi_impl<void, T1, T2, R...> { };


For arithmetic types not subject to promotion, the common type may be viewed as the type of the (possibly mixed-mode) arithmetic expression such as T0() + T1() + ... + Tn().

Defect reports

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

DR Applied to Behavior as published Correct behavior
LWG_2141 C++11 common_type<int, int>::type is int&& decayed result type
LWG_2408 C++11 common_type is not SFINAE-friendly made SFINAE-friendly
LWG_2460 C++11 common_type specializations are nearly impossible to write reduced number of specializations needed


Demonstrates mixed-mode arithmetic on a user-defined class
// Run this code

  #include <iostream>
  #include <type_traits>

  template <class T>
  struct Number { T n; };

  template <class T, class U>
  Number<typename std::common_type<T, U>::type> operator+(const Number<T>& lhs,
                                                          const Number<U>& rhs)
      return {lhs.n + rhs.n};

  int main()
      Number<int> i1 = {1}, i2 = {2};
      Number<double> d1 = {2.3}, d2 = {3.5};
      std::cout << "i1i2: " << (i1 + i2).n << "\ni1d2: " << (i1 + d2).n << '\n'
                << "d1i2: " << (d1 + i2).n << "\nd1d2: " << (d1 + d2).n << '\n';


  i1i2: 3
  i1d2: 4.5
  d1i2: 4.3
  d1d2: 5.8