deductionguidesforstd::packaged_task(3)

NAME
deductionguidesforstd::packaged_task - deductionguidesforstd::packaged_task

Synopsis
   Defined in header <future>
   template< class R, class... Args > (1) (since C++17)
   packaged_task( R(*)(Args...) ) -> packaged_task<R(Args...)>;
   template< class F > (2) (since C++17)
   packaged_task( F ) -> packaged_task</*see below*/>;
   template< class F > (3) (since C++23)
   packaged_task( F ) -> packaged_task</*see below*/>;
   template< class F > (4) (since C++23)
   packaged_task( F ) -> packaged_task</*see below*/>;
   1) This deduction guide is provided for std::packaged_task to allow deduction from
   functions.
   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)>
 }