Clarification needed on UnsafeContinuation documentation

[ole (Ole Begemann)]

@Joe_Groff

They're both right. The task stops executing any async code at all before the continuation is formed, and any state will be moved off of the callstack into the task object at that point. The closure is then immediately executed in the same execution context (in other words, the current thread) with the closure as a parameter. Once the closure returns, control goes back to the executor.

I think this is where my misunderstanding comes from. My mental model is that the task continues running, executing the continuation closure, and that the task suspends when the continuation closure returns. (This begs the question what happens when the continuation closure resumes the continuation synchronously, which would kind of resume the task before it suspended, so maybe this mental model isn't ideal.)

@John_McCall

I think you could argue that the task is still running while the closure is executing. I know that's detectable in at least one way, which is that task-local values are still set; you can't change them without running async code, which you can't do within the closure, but you can still read them. I'm blanking on whether there are other ways this is semantically detectable.

This interpretation better fits my mental model. Another way to detect this seems to be withUnsafeCurrentTask:

func doSomething() async {
  withUnsafeCurrentTask { unsafeTask in
    print("Task before continuation: \(unsafeTask!.hashValue)")
  }
  let _: Void = await withCheckedContinuation { continuation in
    withUnsafeCurrentTask { unsafeTask in
      print("Task inside continuation closure: \(unsafeTask!.hashValue)")
    }
    continuation.resume()
  }
}

This will print the same hash value for both unsafe tasks. (hashValue isn't ideal to verify these are identical, but close enough; another way could be to use withUnsafeCurrentTask to cancel the current task before creating the continuation, then check Task.isCancelled inside the continuation closure.)