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reference_wrapper::operator()
should propagate noexcept
Section: 22.10.6.5 [refwrap.invoke] Status: C++23 Submitter: Zhihao Yuan Opened: 2022-08-31 Last modified: 2023-11-22
Priority: Not Prioritized
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Discussion:
The following assertion does not hold, making the type less capable than the function pointers.
void f() noexcept; std::reference_wrapper fn = f; static_assert(std::is_nothrow_invocable_v<decltype(fn)>);
[2022-09-23; Reflector poll]
Set status to Tentatively Ready after six votes in favour during reflector poll.
Already implemented in MSVC and libstdc++.
[2022-11-12 Approved at November 2022 meeting in Kona. Status changed: Voting → WP.]
Proposed resolution:
This wording is relative to N4910.
Modify 22.10.6.1 [refwrap.general], class template reference_wrapper
synopsis, as indicated:
// 22.10.6.5 [refwrap.invoke], invocation template<class... ArgTypes> constexpr invoke_result_t<T&, ArgTypes...> operator()(ArgTypes&&...) const noexcept(is_nothrow_invocable_v<T&, ArgTypes...>);
Modify 22.10.6.5 [refwrap.invoke] as indicated:
template<class... ArgTypes> constexpr invoke_result_t<T&, ArgTypes...> operator()(ArgTypes&&... args) const noexcept(is_nothrow_invocable_v<T&, ArgTypes...>);-1- Mandates:
-2- Returns:T
is a complete type.INVOKE(get(), std::forward<ArgTypes>(args)...)
. (22.10.4 [func.require])