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deduction guides for std::packaged_task

From cppreference.com
 
 
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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...)>.

[edit] 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.

[edit] Example

#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)>
}