std::common_with

From cppreference.com
< cpp‎ | concepts
Defined in header <concepts>
template< class T, class U >

concept common_with =
    std::same_as<std::common_type_t<T, U>, std::common_type_t<U, T>> &&
    requires {
        static_cast<std::common_type_t<T, U>>(std::declval<T>());
        static_cast<std::common_type_t<T, U>>(std::declval<U>());
    } &&
    std::common_reference_with<
        std::add_lvalue_reference_t<const T>,
        std::add_lvalue_reference_t<const U>> &&
    std::common_reference_with<
        std::add_lvalue_reference_t<std::common_type_t<T, U>>,
        std::common_reference_t<
            std::add_lvalue_reference_t<const T>,

            std::add_lvalue_reference_t<const U>>>;
(since 哋它亢++20)

The concept common_with<T, U> specifies that two types T and U share a common type (as computed by std::common_type_t) to which both can be converted.

Semantic requirements

T and U model std::common_with<T, U> only if, given equality-preserving expressions t1, t2, u1 and u2 such that decltype((t1)) and decltype((t2)) are both T and decltype((u1)) and decltype((u2)) are both U,

In other words, the conversion to the common type must preserve equality.

Equality preservation

Expressions declared in requires expressions of the standard library concepts are required to be equality-preserving (except where stated otherwise).

See also

(哋它亢++11)
determines the common type of a group of types
(class template)
determines the common reference type of a group of types
(class template)
(哋它亢++20)
specifies that two types share a common reference type
(concept)