Can coroutine return std::future? (unable to find the promise type for this coroutine)

I have tried to compile coroutine example from the CppCon presentation https://youtu.be/ZTqHjjm86Bw?t=560

Unfortunately compilation fails:

$ g++-10 -pedantic -Wall -std=c++20 -fcoroutines main.cpp 
main.cpp: In function ‘std::future<int> compute_value()’:
main.cpp:7:16: error: unable to find the promise type for this coroutine
    7 |   int result = co_await std::async([]
      |                ^~~~~~~~

At the beginning presenter warns that what he is about to present is just a proposal. So it makes me confused: can std::future be return from a coroutine, or do I just try to call it incorrectly?

Full code:

#include <coroutine>
#include <iostream>
#include <future>

std::future<int> compute_value(){
  int result = co_await std::async([] 
  {
    return 30;
  });

  co_return result;
}

int main() {
    std::cout << compute_value().get() << std::endl;
}

There are (basically¹) no standard library types in C++20 that implement the necessary coroutine machinery to make coroutines work. This includes std::promise<T>/std::future<T>.

You could write wrappers for them that implement the coroutine machinery though.

¹: There are support types like std::suspend_always/std::suspend_never which have coroutine machinery, but they don't really do anything like what you're thinking of.