template< class... T >
struct common_type;
Determines the common type among all types T...
, that is a type all T...
can be explicitly 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
.
type
.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
.T...
contains exactly two types T1
and T2
),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
;type
denotes that type, see the conditional operator;type
.T...
consists of the types T1, T2, R...
), then if std::common_type<T1, T2>::type exists, the member type
denotes std::common_type<typename std::common_type<T1, T2>::type, R...>::type if such a type exists. In all other cases, there is no member type
.If any type in the parameter pack T
is not a complete type, (possibly cv-qualified) void, or an array of unknown bound, 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.
[edit] Nested types Name Definitiontype
the common type for all T
[edit] Helper types
template< class... T >
using common_type_t = typename common_type<T...>::type;
Users may specialize common_type
for types T1
and T2
if
T1
and T2
depends on a user-defined type, andT1
and T2
.If such a specialization has a member named type
, it must be a public and unambiguous member 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 (except for std::basic_common_reference)(since C++20) from <type_traits>
is undefined.
The following specializations are already provided by the standard library:
[edit] Possible implementation// primary template (used for zero types) template<class...> struct common_type {}; // one type template<class T> struct common_type<T> : common_type<T, T> {}; namespace detail { template<class...> using void_t = void; template<class T1, class T2> using conditional_result_t = decltype(false ? std::declval<T1>() : std::declval<T2>()); template<class, class, class = void> struct decay_conditional_result {}; template<class T1, class T2> struct decay_conditional_result<T1, T2, void_t<conditional_result_t<T1, T2>>> : std::decay<conditional_result_t<T1, T2>> {}; template<class T1, class T2, class = void> struct common_type_2_impl : decay_conditional_result<const T1&, const T2&> {}; // C++11 implementation: // template<class, class, class = void> // struct common_type_2_impl {}; template<class T1, class T2> struct common_type_2_impl<T1, T2, void_t<conditional_result_t<T1, T2>>> : decay_conditional_result<T1, T2> {}; } // two types template<class T1, class T2> struct common_type<T1, T2> : std::conditional<std::is_same<T1, typename std::decay<T1>::type>::value && std::is_same<T2, typename std::decay<T2>::type>::value, detail::common_type_2_impl<T1, T2>, common_type<typename std::decay<T1>::type, typename std::decay<T2>::type>>::type {}; // 3+ types namespace detail { 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<void_t<typename common_type<T1, T2>::type>, T1, T2, R...> : common_type<typename common_type<T1, T2>::type, R...> {}; } template<class T1, class T2, class... R> struct common_type<T1, T2, R...> : detail::common_type_multi_impl<void, T1, T2, R...> {};[edit] Notes
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().
[edit] ExamplesDemonstrates mixed-mode arithmetic on a program-defined class:
#include <iostream> #include <type_traits> template<class T> struct Number { T n; }; template<class T, class U> constexpr Number<std::common_type_t<T, U>> operator+(const Number<T>& lhs, const Number<U>& rhs) { return {lhs.n + rhs.n}; } void describe(const char* expr, const Number<int>& x) { std::cout << expr << " is Number<int>{" << x.n << "}\n"; } void describe(const char* expr, const Number<double>& x) { std::cout << expr << " is Number<double>{" << x.n << "}\n"; } int main() { Number<int> i1 = {1}, i2 = {2}; Number<double> d1 = {2.3}, d2 = {3.5}; describe("i1 + i2", i1 + i2); describe("i1 + d2", i1 + d2); describe("d1 + i2", d1 + i2); describe("d1 + d2", d1 + d2); }
Output:
i1 + i2 is Number<int>{3} i1 + d2 is Number<double>{4.5} d1 + i2 is Number<double>{4.3} d1 + d2 is Number<double>{5.8}[edit] 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 the result type of the conditional operator was not decayed decayed the result type LWG 2408 C++11common_type
was not SFINAE-friendly made SFINAE-friendly LWG 2460 C++11 common_type
specializations were nearly impossible to write reduced the number of
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