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

Section: [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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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 [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> {
        outer-iterator i_ = outer-iterator();               // exposition only
        constexpr explicit value_type(outer-iterator i);    // exposition only
        value_type() = default;
        constexpr explicit value_type(outer-iterator i);
        constexpr inner-iterator<Const> begin() const;
        constexpr default_sentinel_t end() const noexcept;