Task constructor vs Task.Run with async Action - different behavior

Question

Could anyone please explain this, perhaps I'm missing something obvious.

These 2 cases seem to be identical in behavior, and yet they are not.

Case 1:

• Start a Task with an async Action, that does some work for some time:

  var t = Task.Run(async () => { await Task.Delay(2000); });
  

• A second task waits for the first one:

  var waitingTask = Task.Run(() => { t.Wait(); });
  

• Wait for the second task:

  waitingTask.Wait();
  

Case 2:

• Build a Task using the Task constructor, passing the same async Action:

  var t = new Task(async () => { await Task.Delay(2000); });
  

• Start another task to Wait for the first one (just like in the first case):

  var waitingTask = Task.Run(() => { t.Wait(); });
  

• Start the first task:

  t.Start();
  

• Wait for the second task:

  waitingTask.Wait();
  

The first case behaves as expected: the waiting task ends after the first one does, after 2 seconds.

The second case is weird: the waiting task ends very quickly, long before the first one does.

It's easy to see when printing messages from both tasks. A print at the end of the second task will show the difference.

I'm using VS 2015 Preview, which probably uses Roslyn to compile, if this matters.

Answer 1

The answer is here http://blogs.msdn.com/b/pfxteam/archive/2011/10/24/10229468.aspx

In effect the issue is that with new Task, and that it can only take a Action, whilst you are trying to give a Func<Task>. Here are the first two lines again, but rewritten using the correct overloads.

Task t = Task.Run(async () => { await Task.Delay(2000); });

and

Task<Task> t = new Task<Task>(async () => { await Task.Delay(2000); });

The first creates a task as expected. The second creates a task, who's return type is a task that waits 2000ms. The first example works because of the following overloads...

 public Task Run(Func<Task>);
 public Task<T> Run(Func<T>);
 public Task<T> Run(Func<Task<T>>);
 etc...

These overloads are designed to call Task<Task<T>>.Unwrap() automatically for you.

The result is in your second example you are actually awaiting the start/enqueuing of the first task.

You can fix the second example by

var t = new Task<Task>(async () => { await Task.Delay(2000); }).Unwrap();

Answer 2

It's discouraged to use a Task constructor with async-await, you should only be using Task.Run:

var t = Task.Run(async () => await Task.Delay(2000));

The Task constructor was built before the [Task-based Asynchronous Pattern (TAP)][1] and async-await. It will return a task that just starts the inner task and moves on without waiting for it to complete (i.e. fire and forget).

You could solve that by specifying your desired result as Task<Task> and use Unwrap to create a task that represents the entire operation including the inner task:

var t = new Task<Task>(async () => await Task.Delay(2000));   
t.Unwrap().Wait();

Using Task.Run is simpler and does all that for you, including Unwrap so you get back a Task in return that represents the entire async operation.

Answer 3

Answer from Aron is not 100% correct. If you use a fix suggested, you will end up with:

Start may not be called on a promise-style task.

In my case, I solved a similar problem by holding references to both wrapped and unwrapped tasks. You use wrappedTask.Start() to start the task, but unwrappedTask.Wait() to await it (or to access anything related to task's execution, like its TaskStatus).

(Source)