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4135. The helper lambda of std::erase for list should specify return type as bool

Section: 23.3.7.7 [forward.list.erasure], 23.3.11.6 [list.erasure] Status: C++26 Submitter: Hewill Kang Opened: 2024-08-07 Last modified: 2026-06-09

Priority: Not Prioritized

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Discussion:

std::erase for list is specified to return erase_if(c, [&](auto& elem) { return elem == value; }). However, the template parameter Predicate of erase_if only requires that the type of decltype(pred(...)) satisfies boolean-testable, i.e., the return type of elem == value is not necessarily bool.

This means it's worth explicitly specifying the lambda's return type as bool to avoid some pedantic cases (demo): 

#include 

struct Bool {
  Bool(const Bool&) = delete;
  operator bool() const;
};

struct Int {
  Bool& operator==(Int) const;
};

int main() {
  std::list l;
  std::erase(l, Int{}); // unnecessary hard error
}

[2024-08-21; Reflector poll]

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

[Wrocław 2024-11-23; Status changed: Voting → WP.]

Proposed resolution:

This wording is relative to N4988.

  1. Modify 23.3.7.7 [forward.list.erasure] as indicated:

    template
  typename forward_list::size_type
    erase(forward_list& c, const U& value);

    -1- Effects: Equivalent to: return erase_if(c, [&](const auto& elem) -> bool { return elem == value; });

  2. Modify 23.3.11.6 [list.erasure] as indicated:

    template
  typename list::size_type
    erase(list& c, const U& value);

    -1- Effects: Equivalent to: return erase_if(c, [&](const auto& elem) -> bool { return elem == value; });