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std::experimental::ranges::common_type(3) C++ Standard Libary std::experimental::ranges::common_type(3)

NAME

std::experimental::ranges::common_type - std::experimental::ranges::common_type

Synopsis


Defined in header <experimental/ranges/type_traits>
template< class... T > (ranges TS)
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. The
behavior is undefined if any of the types in T... is an incomplete type other than
(possibly cv-qualified) void.


* 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::decay_t<T0>.
* 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
ranges::common_type_t<std::decay_t<T1>, std::decay_t<T2>>, if it
exists; if not, there is no member type.
* Otherwise, (and unless there is a user specialization for
ranges::common_type<T1, T2>), if std::common_type_t<T1, T2> is
well-formed, then the member type denotes that type;
* Otherwise, the member type denotes the type std::decay_t<decltype(false
? std::declval<const T1&>() : std::declval<const T2&>())>, if that
conditional expression is well-formed; if not, there is no member type.


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

Member types


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

Helper types


template< class... T >
using common_type_t = typename common_type<T...>::type;

Specializations


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, ranges::common_type_t<T1, T2>
and ranges::common_type_t<T2, T1> must denote the same type.


A program that adds common_type specializations in violation of these rules has
undefined behavior.

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


This section is incomplete
Reason: no example

See also


common_type determines the common type of a group of types
(C++11) (class template)
common_reference determine the common reference type of a set of types
(class template)

2022.07.31 http://cppreference.com