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

bool (original) (raw)

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

Section: 23.3.7.7 [forward.list.erasure], 23.3.11.6 [list.erasure] Status: WP Submitter: Hewill Kang Opened: 2024-08-07 Last modified: 2024-11-28

Priority: Not Prioritized

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Discussion:

std::erase for list is specified to returnerase_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<class T, class Allocator, class U = T>
   typename forward_list<T, Allocator>::size_type
   erase(forward_list<T, Allocator>& 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<class T, class Allocator, class U = T>
   typename list<T, Allocator>::size_type
   erase(list<T, Allocator>& c, const U& value);

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