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totally_ordered_with both directly and indirectly requires common_reference_withSection: 18.5.5 [concept.totallyordered] Status: C++20 Submitter: United States Opened: 2019-11-07 Last modified: 2021-02-25
Priority: 0
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Discussion:
Addresses US 201
The totally_ordered_with<T, U> redundantly requires both
common_reference_with<const remove_reference_t&, const
remove_reference_t&> and equality_comparable_with<T,
U> (which also has the common_reference_with requirement). The
redundant requirement should be removed.
totally_ordered_with to:
template<class T, class U>
concept totally_ordered_with =
totally_ordered<T> && totally_ordered<U> &&
equality_comparable_with<T, U> &&
totally_ordered<
common_reference_t<
const remove_reference_t<T>&,
const remove_reference_t<U<&>> &&
requires(const remove_reference_t<T<& t,
const remove_reference_t>U>& u) {
[… as before …]
[2019-11 Moved to Ready on Friday AM in Belfast]
Proposed resolution:
This wording is relative to N4835.
Change 18.5.5 [concept.totallyordered] as indicated:
For some type
T, leta,b, andcbe lvalues of typeconst remove_reference_t<T>.Tmodelstotally_orderedonly if
(1.1) — Exactly one of
bool(a < b),bool(a > b), orbool(a == b)istrue.(1.2) — If
bool(a < b)andbool(b < c), thenbool(a < c).(1.3) —
bool(a > b) == bool(b < a).(1.4) —
bool(a <= b) == !bool(b < a).(1.5) —
bool(a >= b) == !bool(a < b).template<class T, class U> concept totally_ordered_with = totally_ordered<T> && totally_ordered<U> &&common_reference_with<const remove_reference_t<T>&, const remove_reference_t<U>&> &&equality_comparable_with<T, U> && totally_ordered< common_reference_t< const remove_reference_t<T>&, const remove_reference_t<U>&>> &&equality_comparable_with<T, U> &&requires(const remove_reference_t<T>& t, const remove_reference_t<U>& u) { { t < u } -> boolean; { t > u } -> boolean; { t <= u } -> boolean; { t >= u } -> boolean; { u < t } -> boolean; { u > t } -> boolean; { u <= t } -> boolean; { u >= t } -> boolean; };