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Section: 25.5.5.4 [common.iter.access] Status: WP Submitter: Jonathan Wakely Opened: 2022-02-12 Last modified: 2022-07-25
Priority: Not Prioritized
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Discussion:
25.5.5.4 [common.iter.access] p5 (5.1) says that common_iterator<T*, S>::operator->() returns std::get<T*>(v_) which has type T* const&. That means that iterator_traits::pointer is T* const& as well (this was recently clarified by LWG 3660). We have an actual pointer here, why are we returning it by reference?
For the three bullet points in 25.5.5.4 [common.iter.access] p5, the second and third return by value anyway, so decltype(auto) is equivalent to auto. For the first bullet, it would make a lot more sense for raw pointers to be returned by value. That leaves the case where the iterator has an operator->() member, which could potentially benefit from returning by reference. But it must return something that is iterator-like or pointer-like, which we usually just pass by value. Casey suggested we should just change common_iterator<I, S>::operator->() to return by value in all cases. Libstdc++ has always returned by value, as an unintended consequence of using a union instead of std::variant<I, S>, so that it doesn't use std::get<I> to return the member.[2022-03-04; Reflector poll]
Set status to Tentatively Ready after six votes in favour during reflector poll.
[2022-07-15; LWG telecon: move to Ready]
[2022-07-25 Approved at July 2022 virtual plenary. Status changed: Ready → WP.]
Proposed resolution:
This wording is relative to N4901.
Modify 25.5.5.1 [common.iterator], class template common_iterator synopsis, as indicated:
[…] constexpr decltype(auto) operator*(); constexpr decltype(auto) operator*() const requires dereferenceable<const I>; constexprdecltype(auto)operator->() const requires see below; […]
Modify 25.5.5.4 [common.iter.access] as indicated:
constexprdecltype(auto)operator->() const requires see below;-3- The expression in the requires-clause is equivalent to: […]
-4- Preconditions: holds_alternative<I>(v_) is true. -5- Effects:
(5.1) — If I is a pointer type or if the expression get<I>(v_).operator->() is well-formed, 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_)); where proxy 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_); } };