std::function
deduction guidesSection: 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
View all other issues in [func.wrap.func.con].
View all issues with C++20 status.
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, ifdecltype(&F::operator())
is of the formR(G::*)(A...) cv &opt noexceptopt
for a class typeG
, then the deduced type isfunction<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Previous resolution [SUPERSEDED]: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 formR(G::*)(A...) cv &opt noexceptopt
for a class typeG
and a sequence of typesA...
. In that case,ifthe deduced type isdecltype(&F::operator())
is of the formR(G::*)(A...) cv &opt noexceptopt
for a class typeG
, thenfunction<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:
-12- Remarks:&F::operator()
is well-formed when treated as an unevaluated operand anddecltype(&F::operator())
is of the formR(G::*)(A...) cv &opt noexceptopt
for a class typeG
.This deduction guide participates in overload resolution only ifThe deduced type is&F::operator()
is well-formed when treated as an unevaluated operand. In that case, ifdecltype(&F::operator())
is of the formR(G::*)(A...) cv &opt noexceptopt
for a class typeG
, then tfunction<R(A...)>
.