deduction guides for std::packaged_task - cppreference.com (original) (raw)

Defined in header
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 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...)>.

  2. 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...)>.

  3. 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   int func(double) { return 0; }   int main() { std::packaged_task f{func}; // deduces packaged_task<int(double)>   int i = 5; std::packaged_task g = & { return i; }; // => packaged_task<int(double)> }