can not await async lambda

84,921

Solution 1

In your lambda example, when you call task.Wait(), you are waiting on the new Task that you constructed, not the delay Task that it returns. To get your desired delay, you would need to also wait on the resulting Task:

Task<Task> task = new Task<Task>(async () => {
    await Task.Delay(1000);
});
task.Start();
task.Wait(); 
task.Result.Wait();

You could avoid constructing a new Task, and just have one Task to deal with instead of two:

Func<Task> task = async () => {
    await TaskEx.Delay(1000);
};
task().Wait();

Solution 2

You need to use TaskEx.RunEx.

It natively supports running async methods on the TaskPool by awaiting the inner task internally. Otherwise you'll run into the issue you're facing, where only the outer task is awaited, which obviously completes immediately, leaving either a task which still needs awaiting, or in your case (and even worse) a void lambda which cannot be awaited.

Alternatively, you can await the task twice, providing you construct your outer task correctly (which currently you are not).

Current code (fixed):

Task task = new Task<Task>(async () =>{
    await TaskEx.Delay(1000);
});

task.Start();
var innerTask = await task;
await innerTask;

Using TaskEx.RunEx:

Task task = TaskEx.RunEx(async () =>{ // Framework awaits your lambda internally.
    await TaskEx.Delay(1000);
});

await task;
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kennyzx
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kennyzx

Gentlemen, it has been a privilege debugging with you tonight. mailto://[email protected] Half the pleasure of coding comes from a good (I mean, really good) keyboard and an ultra-wide monitor.

Updated on July 05, 2022

Comments

  • kennyzx
    kennyzx almost 2 years

    Consider this,

    Task task = new Task (async () =>{
        await TaskEx.Delay(1000);
    });
    task.Start();
    task.Wait(); 
    

    The call task.Wait() does not wait for the task completion and the next line is executed immediately, but if I wrap the async lambda expression into a method call, the code works as expected.

    private static async Task AwaitableMethod()
    {
        await TaskEx.Delay(1000);    
    }
    

    then (updated according comment from svick)

    await AwaitableMethod(); 
    
  • kennyzx
    kennyzx over 11 years
    Nice explanation, but the code TaskEx.Run does not work, still has the same problem.
  • Lawrence Wagerfield
    Lawrence Wagerfield over 11 years
    Argh, sorry! I'm using .NET 4.5... I meant to write TaskEx.RunEx. Compare its signature to TaskEx.Run - you'll see why it's specifically for running async methods.
  • Andrew
    Andrew about 11 years
  • Nelson Rothermel
    Nelson Rothermel over 9 years
    If the first await is after lots of processing you may still want the double tasks. Instead of task.Result.Wait() you can also do task.Unwrap().Wait() (or Unwrap<T>() for non-void methods). The new Task.Run methods automatically unwrap so you only wait on the expected task.
  • drowa
    drowa over 9 years
    As a beginner, I have the impression they could have done a better job with the async keyword; it is very confusing.
  • drowa
    drowa over 9 years
    Is it possible to start the container task without starting the nested task?
  • Scott Chamberlain
    Scott Chamberlain about 9 years
    @drowa async expects functions to return "Hot" tasks to be returned from functions that are already started. If you await a task that has not been started, and will not be started, the program will never come back from the await. The only way to "not start the nested task" is while in the container function to never call the inner function that creates the nested task.
  • VoteCoffee
    VoteCoffee over 4 years
    The potential pitfalls link is broken and is now devblogs.microsoft.com/pfxteam/…