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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 C++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.

[edit] 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.

[edit] Equality preservation

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

[edit] See also

determines the common type of a group of types
(class template) [edit]
determines the common reference type of a group of types
(class template) [edit]
specifies that two types share a common reference type
(concept) [edit]