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)>
}