获取 std::future 的状态

Question

Is it possible to check if a std::future has finished or not? As far as I can tell the only way to do it would be to call wait_for with a zero duration and check if the status is ready or not, but is there a better way?

Answer 1

You are correct, and apart from calling wait_until with a time in the past (which is equivalent) there is no better way.

You could always write a little wrapper if you want a more convenient syntax:

template<typename R>
  bool is_ready(std::future<R> const& f)
  { return f.wait_for(std::chrono::seconds(0)) == std::future_status::ready; }

NB if the function is deferred this will never return true, so it's probably better to check wait_for directly in the case where you might want to run the deferred task synchronously after a certain time has passed or when system load is low.

Answer 2

There's an is_ready member function in the works for std::future. In the meantime, the VC implementation has an _Is_ready() member.

Answer 3

My first bet would be to call wait_for with a 0 duration, and check the result code that can be one of future_status::ready , future_status::deferred or future_status::timeout .

valid() will return true if *this refers to a shared state, independently of whether that state is ready or not. See cppreference .