deduction guides for std::packaged_task

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std::packaged_task
Defined in header <future>
template < class R, class ... Args >
packaged_task ( R ( * ) ( Args... ) ) - > packaged_task < R ( Args... ) > ;
(1) (since C++17)
template < class F >
packaged_task ( F ) - > packaged_task < /*see below*/ > ;
(2) (since C++17)
template < class F >
packaged_task ( F ) - > packaged_task < /*see below*/ > ;
(3) (since C++23)
template < class F >
packaged_task ( F ) - > packaged_task < /*see below*/ > ;
(4) (since C++23)
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)>
}
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