deduction guides for 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) |
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) for some class type G. The deduced type is
std::packaged_task<R(A...)>
.
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; }; // deduces packaged_task<int(double)> }
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