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4135. The helper lambda ofstd::erase for list should specify return type as bool
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
View all issues with WP status.
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.
bool to avoid some pedantic cases (demo):
#include <list>
struct Bool {
Bool(const Bool&) = delete;
operator bool() const;
};
struct Int {
Bool& operator==(Int) const;
};
int main() {
std::list<Int> 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.
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; });
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; });
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