C# async function as Task<int>

Question

In the following code:

async static Task<int> test(){ 
    Console.WriteLine("{0}: Start 3", DateTime.Now);
    await Task.Delay(3000);
    return 3; 
}

List<Task<int>> tasks2 = new List<Task<int>> {
    test(),
    new Task<int>( delegate { Console.WriteLine("{0}: Start 3", DateTime.Now); Task.Delay(3000).Wait(); return 3; } ),
    new Task<int>( delegate { Console.WriteLine("{0}: Start 1", DateTime.Now); Task.Delay(1000).Wait(); return 1; } ),
};


foreach (var task in tasks2)
    task.Start(); // this cause exception at runtime

Why I can't use test method as Task<int>? If function returns Task<int>...

Yacoub Massad · Accepted Answer · 2016-01-20 20:49:28Z

3

The task returned by test() has already started when you invoked test(). So you cannot start it again.

Actually, the task returned by test() is a promise-style task that does not run on the thread-pool thread in the first place.

You should almost always create tasks that have already started (once they are created).

I.e., you should not use the constructor of Task and then call the Start method.

Instead, if you want to run the task on a thread-pool thread, use Task.Run.

edited Jan 20, 2016 at 20:49

answered Jan 20, 2016 at 20:42

Yacoub Massad

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4 Comments

tower120 Over a year ago

"You should almost always create tasks that have already started (once they are created)." - Why? What if I want to run them latter?

tower120 Over a year ago

" the task returned by test() is a promise-style task" - What is this?

Yacoub Massad Over a year ago

Because Task.Run is a better alternative. Take a look at this article. If you want to create something and start it later, use Action<T>, and then start it via Task.Run later.

Yacoub Massad Over a year ago

Here is a good article about promise tasks.

Martin · Accepted Answer · 2016-01-20 22:02:15Z

0

As others mentioned you cannot call Start as it was already started. You could still achieve your goals by checking for the status, if that is what you wish for. Please see the updated code below:

async static Task<int> test(){ 
    Console.WriteLine("{0}: Start 3", DateTime.Now);
    await Task.Delay(3000);
    return 3; 
}


        List<Task<int>> tasks2 = new List<Task<int>> {
           new Task<int>( delegate{ return test().Result;}),
            new Task<int>( delegate { Console.WriteLine("{0}: Start 3", DateTime.Now); Task.Delay(3000).Wait(); return 3; } ),
            new Task<int>( delegate { Console.WriteLine("{0}: Start 1", DateTime.Now); Task.Delay(1000).Wait(); return 1; } ),
            };

foreach (var task in tasks2)
{
        task.Start();
}

edited Jan 20, 2016 at 22:02

answered Jan 20, 2016 at 20:52

Martin

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2 Comments

tower120 Over a year ago

I agree, it already started. Is there a way, to pass test function as Task<int> to the List<Task<int>> ?

Martin Over a year ago

I have updated the List initialization part. You could now call Start when you require.

WiSeeker · Accepted Answer · 2016-01-20 21:06:48Z

Like everyone else has pointed out, the task has been set in motion when you called test() so you can't start on it again. Give below code a shot to see what's going on.

As you can see, the console output comes for each of the task. Task returned by test() is started by calling that method. So, this code only explicitly starts the other tasks that are in Created status. Finally, we wait for all tasks to complete by calling WhenAll() which in turn returns a task on which we wait. You will notice the longest delay of 3 seconds (of all tasks) before "All done!" is printed.

public static class AsyncTest
{
    static async Task<int> test()
    {
        Console.WriteLine("{0}: Start 3", DateTime.Now);
        await Task.Delay(3000);
        return 3;
    }

    public static void Main()
    {

        var tasks2 = new List<Task<int>>
        {
            test(),
            new Task<int>( delegate { Console.WriteLine("{0}: Start 3", DateTime.Now); Task.Delay(3000).Wait(); return 3; } ),
            new Task<int>( delegate { Console.WriteLine("{0}: Start 1", DateTime.Now); Task.Delay(1000).Wait(); return 1; } )
        };

        foreach (var task in tasks2)
        {
            if (task.Status == TaskStatus.Created)
                task.Start();
        }

        Task.WhenAll(tasks2).Wait();

        Console.WriteLine("All done!");
    }
}

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