not_fn call_wrapper
can form invalid typesSection: 22.10.13 [func.not.fn] Status: C++17 Submitter: Jonathan Wakely Opened: 2016-08-19 Last modified: 2021-06-06
Priority: 0
View all other issues in [func.not.fn].
View all issues with C++17 status.
Discussion:
The definition of the call_wrapper
type in the C++17 CD means this
fails to compile:
#include <functional> struct abc { virtual void f() const = 0; }; struct derived : abc { void f() const { } }; struct F { bool operator()(abc&) { return false; } }; derived d; bool b = std::not_fn(F{})(static_cast<abc&&>(d));
The problem is that the return types use result_of_t<F(abc)>
and
F(abc)
is not a valid function type, because it takes an abstract
class by value.
result_of_t<F(Args&&...)>
instead.
[2016-09-09 Issues Resolution Telecon]
P0; move to Tentatively Ready
Proposed resolution:
This wording is relative to N4606.
Modify [func.not_fn], class call_wrapper
synopsis, as indicated:
class call_wrapper { […] template<class... Args> auto operator()(Args&&...) & -> decltype(!declval<result_of_t<FD&(Args&&...)>>()); template<class... Args> auto operator()(Args&&...) const& -> decltype(!declval<result_of_t<FD const&(Args&&...)>>()); template<class... Args> auto operator()(Args&&...) && -> decltype(!declval<result_of_t<FD(Args&&...)>>()); template<class... Args> auto operator()(Args&&...) const&& -> decltype(!declval<result_of_t<FD const(Args&&...)>>()); […] };
Modify the prototype declarations of [func.not_fn] as indicated:
template<class... Args> auto operator()(Args&&... args) & -> decltype(!declval<result_of_t<FD&(Args&&...)>>()); template<class... Args> auto operator()(Args&&... args) const& -> decltype(!declval<result_of_t<FD const&(Args&&...)>>());[…]
template<class... Args> auto operator()(Args&&... args) && -> decltype(!declval<result_of_t<FD(Args&&...)>>()); template<class... Args> auto operator()(Args&&... args) const&& -> decltype(!declval<result_of_t<FD const(Args&&...)>>());