Get the status of a std::future
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; }N.B. 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.
There's an is_ready member function in the works for std::future. In the meantime, the VC implementation has an _Is_ready() member.
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.