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3238. Insufficiently-defined behavior of std::function deduction guides

Section: 20.14.16.2.1 [func.wrap.func.con] Status: WP Submitter: Louis Dionne Opened: 2019-07-17 Last modified: 2020-09-06

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 20.14.16.2.1 [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Previous resolution [SUPERSEDED]: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.

  1. Modify 20.14.16.2.1 [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, then the 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.

  1. Modify 20.14.16.2.1 [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...)>.