reverse_iterator
/common_iterator
's operator->
should not require the underlying iterator's operator->
to be a const
member functionSection: 24.5.1.6 [reverse.iter.elem], 24.5.5.4 [common.iter.access] Status: NAD Submitter: Hewill Kang Opened: 2022-06-27 Last modified: 2022-11-30
Priority: Not Prioritized
View all other issues in [reverse.iter.elem].
View all issues with NAD status.
Discussion:
For non-pointer types, reverse_iterator::operator->
requires that the Iterator
must have an operator->()
with const
-qualifier, whereas in the Effects clause,
it always invokes the non-const
object's operator->()
.
common_iterator::operator->
also requires that I::operator->()
must be
const
-qualified, which seems reasonable since the return type of get<I>(v_)
is const I&
. However, LWG 3672 makes common_iterator::operator->()
always return a value, which makes it unnecessary to detect the constness of I::operator->()
,
because it will be invoked with a non-const
returned object anyway.
I think we should remove this constraint as I don't see the benefit of doing that.
Constraining iterator's operator->()
to be const
and finally invoking non-const
overload doesn't feel right to me either. In <ranges>
, the exposition-only constraint
has-arrow
(25.5.2 [range.utility.helpers]) for operator->()
does not
require that the underlying iterator's operator->()
to be const
, we should make them
consistent, and I believe this relaxation of constraints can bring some value.
Daniel:
This issue's second part of the resolution actually depends on 3672 being applied. But
note that the reference wording below is still N4910.
[2022-08-23; Reflector poll: NAD]
Implicit variations apply to those requires-expressions, so calling as non-const (and rvalue) is fine. The PR actually loses that property and makes those overloads truly underconstrained. Motivation for relaxing it is vague. As for consistency, we should fix has-arrow instead.
[2022-11-30 LWG telecon. Status changed: Tentatively NAD → NAD.]
Proposed resolution:
This wording is relative to N4910.
Modify 24.5.1.6 [reverse.iter.elem] as indicated:
constexpr pointer operator->() const requires (is_pointer_v<Iterator> || requires(constIterator i) { i.operator->(); });-2- Effects:
(2.1) — If
Iterator
is a pointer type, equivalent to:return prev(current);
(2.2) — Otherwise, equivalent to:
return prev(current).operator->();
Modify 24.5.5.4 [common.iter.access] as indicated:
constexpr decltype(auto) operator->() const requires see below;-3- The expression in the requires-clause is equivalent to:
indirectly_readable<const I> && (requires(constI&i) { i.operator->(); } || is_reference_v<iter_reference_t<I>> || constructible_from<iter_value_t<I>, iter_reference_t<I>>)-4- Preconditions:
-5- Effects:holds_alternative<I>(v_)
istrue
.
(5.1) — If
I
is a pointer type or ifthe expressionget<I>(v_).operator->()
is well-formedrequires(I i) { i.operator->(); }
istrue
, equivalent to:return get<I>(v_);
(5.2) — Otherwise, if
iter_reference_t<I>
is a reference type, equivalent to:auto&& tmp = *get<I>(v_); return addressof(tmp);(5.3) — Otherwise, equivalent to:
return proxy(*get<I>(v_));
whereproxy
is the exposition-only class:class proxy { iter_value_t<I> keep_; constexpr proxy(iter_reference_t<I>&& x) : keep_(std::move(x)) {} public: constexpr const iter_value_t<I>* operator->() const noexcept { return addressof(keep_); } };[…]