std::common_type

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

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, see the conditional operator;

Otherwise, if std::decay<decltype(false ? std::declval<CR1>() : std::declval<CR2>())>::type is a valid type, where CR1 and CR2 are const std::remove_reference_t<T1>& and const std::remove_reference_t<T2>& respectively, 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<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.

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:

std::common_type<std::chrono::duration> (C++11)	specializes the std::common_type trait (class template specialization) [edit]
std::common_type<std::chrono::time_point> (C++11)	specializes the std::common_type trait (class template specialization) [edit]
std::common_type<std::pair> (C++23)	determines the common type of two `pair`s (class template specialization) [edit]
std::common_type<tuple-like> (C++23)	determines the common type of a `tuple` and a `tuple-like` type (class template specialization) [edit]
std::common_type<std::basic_const_iterator> (C++23)	determines the common type of an iterator and an adapted `basic_const_iterator` type (class template specialization) [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...> {};

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().

Examples

Demonstrates mixed-mode arithmetic on a program-defined class:

Run this code

#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}

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++11	`common_type` was not SFINAE-friendly	made SFINAE-friendly
LWG 2460	C++11	`common_type` specializations were nearly impossible to write	reduced the number of specializations needed

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

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std::common_type

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