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Section: 22.10.17.3.2 [func.wrap.func.con] Status: C++20 Submitter: Louis Dionne Opened: 2019-07-17 Last modified: 2021-02-25
Priority: Not Prioritized
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Discussion:
The following code is currently undefined behavior:
#include <functional> struct R { }; struct f0 { R operator()() && { return {}; } }; int main() { std::function f = f0{}; }
The reason is that 22.10.17.3.2 [func.wrap.func.con]/12 says:
This deduction guide participates in overload resolution only if &F::operator() is well-formed when treated as an unevaluated operand. In that case, if decltype(&F::operator()) is of the form R(G::*)(A...) cv &opt noexceptopt for a class type G, then the deduced type is function<R(A...)>.
However, it does not define the behavior when &F::operator() is well-formed but not of the required form (in the above example it's of the form R(G::*)(A...) &&, which is rvalue-reference qualified instead of optionally-lvalue-reference qualified). libc++'s implementation of the deduction guide SFINAE's out when either &F::operator() is not well-formed, or it is not of the required form. It seems like mandating that behavior in the Standard is the way to go.
Previous resolution [SUPERSEDED]:
This wording is relative to N4820.
Modify 22.10.17.3.2 [func.wrap.func.con] as indicated:
template<class F> function(F) -> function<see below>;-12- Remarks: This deduction guide participates in overload resolution only if &F::operator() is well-formed when treated as an unevaluated operand, and its type is of the form R(G::*)(A...) cv &opt noexceptopt for a class type G and a sequence of types A.... In that case,
if decltype(&F::operator()) is of the form R(G::*)(A...) cv &opt noexceptopt for a class type G, thenthe deduced type is function<R(A...)>.
[2020-02-13; Prague]
LWG improves wording matching Marshall's Mandating paper.
[Status to Immediate on Friday in Prague.]
Proposed resolution:
This wording is relative to N4849.
Modify 22.10.17.3.2 [func.wrap.func.con] as indicated:
[Drafting note: This edit should be used instead of the corresponding edit in P1460]
template<class F> function(F) -> function<see below>;-?- Constraints: &F::operator() is well-formed when treated as an unevaluated operand and decltype(&F::operator()) is of the form R(G::*)(A...) cv &opt noexceptopt for a class type G.
-12- Remarks:This deduction guide participates in overload resolution only if &F::operator() is well-formed when treated as an unevaluated operand. In that case, if decltype(&F::operator()) is of the form R(G::*)(A...) cv &opt noexceptopt for a class type G, then tThe deduced type is function<R(A...)>.