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3755. tuple-for-each can call user-defined operator,

Section: 26.7.24.2 [range.zip.view] Status: WP Submitter: Nicole Mazzuca Opened: 2022-08-26 Last modified: 2022-11-17

Priority: Not Prioritized

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Discussion:

The specification for tuple-for-each is:

template<class F, class Tuple>
constexpr auto tuple-for-each(F&& f, Tuple&& t) { // exposition only
  apply([&]<class... Ts>(Ts&&... elements) {
    (invoke(f, std::forward<Ts>(elements)), ...);
  }, std::forward<Tuple>(t));
}

Given

struct Evil {
  void operator,(Evil) {
        abort();
    }
};

and tuple<int, int> t, then tuple-for-each([](int) { return Evil{}; }, t), the program will (unintentionally) abort.

It seems likely that our Evil's operator, should not be called.

[2022-09-23; Reflector poll]

Set status to Tentatively Ready after nine votes in favour during reflector poll.

Feedback from reviewers:

"NAD. This exposition-only facility is only used with things that return void. As far as I know, users can't define operator, for void." "If I see the void cast, I don't need to audit the uses or be concerned that we'll add a broken use in the future."

[2022-11-12 Approved at November 2022 meeting in Kona. Status changed: Voting → WP.]

Proposed resolution:

This wording is relative to the forthcoming C++23 CD.

• Modify [range.adaptor.tuple] as indicated:

  template<class F, class Tuple>
  constexpr auto tuple-for-each(F&& f, Tuple&& t) { // exposition only
    apply([&]<class... Ts>(Ts&&... elements) {
      (static_cast<void>(invoke(f, std::forward<Ts>(elements))), ...);
    }, std::forward<Tuple>(t));
  }