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Section: 22.2.6 [declval] Status: New Submitter: Jiang An Opened: 2023-03-07 Last modified: 2023-03-22
Priority: 4
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Discussion:
Currently libc++ and libstdc++ determine the return type of std::declval like this:
template<class _Tp> _Tp&& __declval_ret(int); // selected when _Tp is a referenceable type template<class _Tp> _Tp __declval_ret(long); // selected when _Tp is cv void template<class _Tp> decltype(__declval_ret<_Tp>(0)) declval() noexcept;
This strategy avoids instantiation of class templates. But it also drops cv-qualifiers of the return type when the type is cv void, which is different from the standard requirements. Such difference has no impact in normal use of std::declval, but is observable via decltype(std::declval<const void>) and its friends.
Given maintainers may think it's reasonable to keep the current implementation (see GCC Bugzilla #109049), it may be worthwhile to legitimate such strategy. Should we make such construction ill-formed?[2023-03-22; Reflector poll]
Set priority to 4 after reflector poll.
"The testcase isn't even valid with the previous 'conforming' libstdc++ implementation."
"declval isn't an addressable function, so would prefer if this
was ill-formed rather than complicating the definition for this case."
Proposed resolution:
This wording is relative to N4928.
Modify 22.2.1 [utility.syn], header <utility> synopsis, and 22.2.6 [declval] as indicated:
template<class T> remove_cv_t<add_rvalue_reference_t<T>> declval() noexcept; // as unevaluated operand