deduction guides for std::packaged_task
From cppreference.com
2) This overload participates in overload resolution only if
&F::operator() is well-formed when treated as an unevaluated operand and decltype(&F::operator()) is of the form R(G::*)(A...) (optionally cv-qualified, optionally noexcept, optionally lvalue reference qualified). The deduced type is std::packaged_task<R(A...)>.3) This overload participates in overload resolution only if
&F::operator() is well-formed when treated as an unevaluated operand and F::operator() is an explicit object parameter function whose type is of form R(G, A...) or R(G, A...) noexcept. The deduced type is std::packaged_task<R(A...)>.4) This overload participates in overload resolution only if
&F::operator() is well-formed when treated as an unevaluated operand and F::operator() is a static member function whose type is of form R(A...) or R(A...) noexcept. The deduced type is std::packaged_task<R(A...)>.Notes
These deduction guides do not allow deduction from a function with ellipsis parameter, and the ... in the types is always treated as a pack expansion.
Example
Run this code
#include <future>
int func(double) { return 0; }
int main()
{
std::packaged_task f{func}; // deduces packaged_task<int(double)>
int i = 5;
std::packaged_task g = [&](double) { return i; }; // => packaged_task<int(double)>
}