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3727. reverse_iterator/common_iterator's operator-> should not require the underlying iterator's operator-> to be a const member function

Section: 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].

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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⁽ⁱ⁾ 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⁽ⁱ⁾ 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.

1. Modify 24.5.1.6 [reverse.iter.elem] as indicated:

   constexpr pointer operator->() const
     requires (is_pointer_v<Iterator> ||
               requires(const Iterator i) { i.operator->(); });
   

   -2- Effects:

   1. (2.1) — If Iterator is a pointer type, equivalent to: return prev(current);

   2. (2.2) — Otherwise, equivalent to: return prev(current).operator->();

2. 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(const I& i) { i.operator->(); } ||
     is_reference_v<iter_reference_t<I>> ||
     constructible_from<iter_value_t<I>, iter_reference_t<I>>)
     

   -4- Preconditions: holds_alternative<I>(v_) is true.

   -5- Effects:

   1. (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_);

   2. (5.2) — Otherwise, if iter_reference_t<I> is a reference type, equivalent to:

        auto&& tmp = *get<I>(v_);
        return addressof(tmp);
        

   3. (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_);
          }
        };
        

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