4013. lazy_split_view::outer-iterator::value_type should not provide default constructor

Section: 26.7.16.4 [range.lazy.split.outer.value] Status: New Submitter: Hewill Kang Opened: 2023-11-11 Last modified: 2023-11-18 13:05:11 UTC

Priority: Not Prioritized

View all other issues in [range.lazy.split.outer.value].

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Discussion:

After P2325, there is no reason for lazy_split_view::outer-iterator::value_type to provide a default constructor, which only leads to unexpected behavior:

#include <ranges>

constexpr int arr[] = {42};
constexpr auto split = arr | std::views::lazy_split(0);
static_assert(!std::ranges::range_value_t<decltype(split)>{});  // UB, dereferencing a null pointer

Also, the other constructor should be private because it makes no sense for the user to construct it arbitrarily, which is not the intention.

Proposed resolution:

This wording is relative to N4964.

1. Modify 26.7.16.4 [range.lazy.split.outer.value], class split_view::outer-iterator::value_type synopsis, as indicated:

   namespace std::ranges {
     template<input_range V, forward_range Pattern>
       requires view<V> && view<Pattern> &&
                indirectly_comparable<iterator_t<V>, iterator_t<Pattern>, ranges::equal_to> &&
                (forward_range<V> || tiny-range<Pattern>)
     template<bool Const>
     struct lazy_split_view<V, Pattern>::outer-iterator<Const>::value_type
       : view_interface<value_type> {
     private:
       outer-iterator i_ = outer-iterator();               // exposition only
     
       constexpr explicit value_type(outer-iterator i);    // exposition only
   
     public:
       value_type() = default;
       constexpr explicit value_type(outer-iterator i);
   
       constexpr inner-iterator<Const> begin() const;
       constexpr default_sentinel_t end() const noexcept;
     };
   }