Why the following program never ends?
namespace Example
{
class Program
{
static async Task Main(string[] args)
{
var result = await new Program().test();
Console.WriteLine(result);
}
private async Task<(int, string)> test()
{
var result = new Task<(int, string)>(() => (10, "sssss"));
return (result.Result.Item1, result.Result.Item2);
}
}
}
What I have noticed is that result.Result or writing await new Task<(int, string)>(() => (10, "sssss")); will let the program keep awaiting forever!!
The Task constructor creates a "cold" task, a task that has not been started. If you await such a task, the await will never complete, because the task will remain forever in the Created status, and will never transition to the RanToCompletion status. To start a task that was created cold, you must call its Start or RunSynchronously methods, preferably passing the TaskScheduler.Default as argument:
private Task<(int, string)> TestAsync()
{
var task = new Task<(int, string)>(() => (10, "sssss"));
task.Start(TaskScheduler.Default);
return task;
}
Creating cold tasks by using the Task constructor is an advanced technique that is used rarely in practice. The common way to create delegate-based tasks is by using the static Task.Run method, that creates hot tasks. The method below is functionally identical¹ with the previous method:
private Task<(int, string)> TestAsync()
{
return Task.Run(() => (10, "sssss"));
}
Be aware that both methods above are considered bad practices, for reasons explained here.
¹ Actually not exactly identical. The second example starts a task having the TaskCreationOptions.DenyChildAttach configuration. More info about this can be found here.
Mixing async await and blocking calls like .Result can causes deadlocks
No need to await if nothing in there is async. Just return Task.FromResult with your desired value
private Task<(int, string)> test() {
(int, string) value = (10, "sssss");
return Task.FromResult(value);
}
It will return a successfully completed task with the specified result.
Related
I know I may be downvoted but apparently I just don't understand async-await enough and the questions/answer I found, and the articles I found, didn't help me to find an answer for this question:
How can I make "2" to be printed out? Or actually, WHY doesn't 2 gets printed out, both in await t and in t.Wait() ?:
static Task t;
public static async void Main()
{
Console.WriteLine("Hello World");
t = GenerateTask();
await t;
//t.Wait();
Console.WriteLine("Finished");
}
public static Task GenerateTask()
{
var res = new Task(async () =>
{
Console.WriteLine("1");
await Task.Delay(10000);
Console.WriteLine("2");
});
res.Start();
return res;
}
Edit: I'm creating a task and returning it cause in real-life I need to await on this task later on, from a different method.
Edit2: await Task.Delay is just a placeholder for a real-life await on a different function.
Printing '2'
The 2 is actually printed, 10 seconds after 1 is printed. You can observe this if you add Console.ReadLine(); after printing 'Finished'.
The output is
Hello World
1
Finished
2
What is happening?
When you await t (which is res in GenerateTask method) you are awaiting the created Task and not the task that res created.
How to fix (fancy way)
You will need to await both the outer task and inner task. To be able to await the inner task you need to expose it. To expose it you need to change the type of the task from Task to Task<Task> and the return type from Task to Task<Task>.
It could look something like this:
public static async Task Main()
{
Console.WriteLine("Hello World");
var outerTask = GenerateTask();
var innerTask = await outerTask; // what you have
await innerTask; // extra await
Console.WriteLine("Finished");
Console.ReadLine();
}
public static Task<Task> GenerateTask() // returns Task<Task>, not Task
{
var res = new Task<Task>(async () => // creates Task<Task>, not Task
{
Console.WriteLine("1");
await Task.Delay(TimeSpan.FromSeconds(10));
Console.WriteLine("2");
});
res.Start();
return res;
}
The output now is:
Hello World
1
2
Finished
How to fix (easy way)
The outer task is not needed.
public static async Task Main()
{
Console.WriteLine("Hello World");
var t = GenerateTask();
await t;
Console.WriteLine("Finished");
Console.ReadLine();
}
public static async Task GenerateTask()
{
Console.WriteLine("1");
await Task.Delay(TimeSpan.FromSeconds(10));
Console.WriteLine("2");
}
It looks like it's because the constructor to new Task only takes some form of an Action (So the Task never gets returned even though it's async). So essentially what you're doing is an Async void with your delegate. Your await Task.Delay(10000) is returning and the action is considered 'done'.
You can see this if you change the await Task.Delay(10000) to Task.Delay(10000).Wait() and remove the async from the delegate.
On another note though, I've never personally seen or used new Task before. Task.Run() is a much more standard way to do it, and it'll allow for the await to be used. Also means you don't have to call Start() yourself.
Also you might already know this but, in this specific case you don't need a new task at all. You can just do this:
public static async Task GenerateTask()
{
Console.WriteLine("1");
await Task.Delay(10000);
Console.WriteLine("2");
}
Regarding your edits
Replacing your GenerateTask with what I wrote should do what you want. The async/await will turn your method into a Task that has started execution. This is exactly what you are trying to do so I'm not quite sure what you are asking with your edits.
The task returned from GenerateTask can be awaited whenever you want, or not awaited at all. You should almost never need to do new Task(). The only reason I can think is if you wanted to delay execution of the task until later, but there would be better ways around it rather than calling new Task().
If you use the way I showed in your real-life situation, let me know what doesn't work about it and I'll be happy to help.
You should use Task.Run() rather than creating a Task directly:
public static Task GenerateTask()
{
return Task.Run(async () =>
{
Console.WriteLine("1");
await Task.Delay(10000);
Console.WriteLine("2");
});
}
Task.Start() doesn't work because it doesn't understand async delegates, and the returned task just represents the beginning of the task.
Note that you can't fix this by using Task.Factory.StartNew() either, for the same reason.
See Stephen Cleary's blog post on this issue, from which I quote:
[Task.Factory.StartNew()] Does not understand async delegates. This is actually the same as
point 1 in the reasons why you would want to use StartNew. The problem
is that when you pass an async delegate to StartNew, it’s natural to
assume that the returned task represents that delegate. However, since
StartNew does not understand async delegates, what that task actually
represents is just the beginning of that delegate. This is one of the
first pitfalls that coders encounter when using StartNew in async
code.
These comments also apply to the Task constructor, which also doesn't understand async delegates.
However, it's important to note that if you are already awaiting in the code and you don't need to parallelise some compute-bound code, you don't need to create a new task at all - just using the code in your Task.Run() on its own will do.
In an answer to one of my other questions, I was told that use of new Task(() => { }) is not something that is a normal use case. I was advised to use Func<Task> instead. I have tried to make that work, but I can't seem to figure it out. (Rather than drag it out in the comments, I am asking a separate question here.)
My specific scenario is that I need the Task to not start right when it is declared and to be able to wait for it later.
Here is a LinqPad example using new Task(() => { }). NOTE: This works perfectly! (Except that it uses new Task.)
static async void Main(string[] args)
{
// Line that I need to swap to a Func<Task> somehow.
// note that this is "cold" not started task
Task startupDone = new Task(() => { });
var runTask = DoStuff(() =>
{
//+++ This is where we want to task to "start"
startupDone.Start();
});
//+++ Here we wait for the task to possibly start and finish. Or timeout.
// Note that this times out at 1000ms even if "blocking = 10000" below.
var didStartup = startupDone.Wait(1000);
Console.WriteLine(!didStartup ? "Startup Timed Out" : "Startup Finished");
await runTask;
Console.Read();
}
public static async Task DoStuff(Action action)
{
// Swap to 1000 to simulate starting up blocking
var blocking = 1; //1000;
await Task.Delay(500 + blocking);
action();
// Do the rest of the stuff...
await Task.Delay(1000);
}
I tried swapping the second line with:
Func<Task> startupDone = new Func<Task>(async () => { });
But then the lines below the comments with +++ in them don't work right.
I swapped the startupDone.Start() with startupDone.Invoke().
But startupDone.Wait needs the task. Which is only returned in the lambda. I am not sure how to get access to the task outside the lambda so I can Wait for it.
How can use a Func<Task> and start it in one part of my code and do a Wait for it in another part of my code? (Like I can with new Task(() => { })).
The code you posted cannot be refactored to make use of a Func<Task> instead of a cold task, because the method that needs to await the task (the Main method) is not the same method that controls the creation/starting of the task (the lambda parameter of the DoStuff method). This could make the use of the Task constructor legitimate in this case, depending on whether the design decision to delegate the starting of the task to a lambda is justified. In this particular example the startupDone is used as a synchronization primitive, to signal that a condition has been met and the program can continue. This could be achieved equally well by using a specialized synchronization primitive, like for example a SemaphoreSlim:
static async Task Main(string[] args)
{
var startupSemaphore = new SemaphoreSlim(0);
Task runTask = RunAsync(startupSemaphore);
bool startupFinished = await startupSemaphore.WaitAsync(1000);
Console.WriteLine(startupFinished ? "Startup Finished" : "Startup Timed Out");
await runTask;
}
public static async Task RunAsync(SemaphoreSlim startupSemaphore)
{
await Task.Delay(500);
startupSemaphore.Release(); // Signal that the startup is done
await Task.Delay(1000);
}
In my opinion using a SemaphoreSlim is more meaningful in this case, and makes the intent of the code clearer. It also allows to await asynchronously the signal with a timeout WaitAsync(Int32), which is not something that you get from a Task out of the box (it is doable though).
Using cold tasks may be tempting in some cases, but when you revisit your code after a month or two you'll find yourself confused, because of how rare and unexpected is to have to deal with tasks that may or may have not been started yet.
I always try my hardest to never have blocking behavior when dealing with anything async or any type that represents potential async behavior such as Task. You can slightly modify your DoStuff to facilitate waiting on your Action.
static async void Main(string[] args)
{
Func<CancellationToken,Task> startupTask = async(token)=>
{
Console.WriteLine("Waiting");
await Task.Delay(3000, token);
Console.WriteLine("Completed");
};
using var source = new CancellationTokenSource(2000);
var runTask = DoStuff(() => startupTask(source.Token), source.Token);
var didStartup = await runTask;
Console.WriteLine(!didStartup ? "Startup Timed Out" : "Startup Finished");
Console.Read();
}
public static async Task<bool> DoStuff(Func<Task> action, CancellationToken token)
{
var blocking = 10000;
try
{
await Task.Delay(500 + blocking, token);
await action();
}
catch(TaskCanceledException ex)
{
return false;
}
await Task.Delay(1000);
return true;
}
First, the type of your "do this later" object is going to become Func<Task>. Then, when the task is started (by invoking the function), you get back a Task that represents the operation:
static async void Main(string[] args)
{
Func<Task> startupDoneDelegate = async () => { };
Task startupDoneTask = null;
var runTask = await DoStuff(() =>
{
startupDoneTask = startupDoneDelegate();
});
var didStartup = startupDoneTask.Wait(1000);
Console.WriteLine(!didStartup ? "Startup Timed Out" : "Startup Finished");
}
Typically, I do the following
public static async Task dosth()
{
List<Task> job = new List<Task>();
for (int i = 0; i < 3; i++)
{
job.Add(sleep());
}
Task.WhenAll(job.ToArray());
}
static async Task sleep()
{
await Task.Delay(1000);
Console.WriteLine("Finish new");
}
It works smoothly, no problem. But when I do a review on my own code (trying using other syntax to do the same job), I suddenly figure out the following two are different.
public static async Task dosthA()
{
//This will be working synchronously, take 3 seconds.
await sleep();
await sleep();
await sleep();
//This will be working asynchronously, take 1 second only.
Task A = sleep();
Task B = sleep();
Task C = sleep();
await A;
await B;
await C;
}
Why assigning the async function to a new variable make difference? I originally think they are the same.
Update
Why it is confusing me is, actually in Microsoft doc on Async-await,
They stated the following in their code.
// Calls to TaskOfTResult_MethodAsync
Task<int> returnedTaskTResult = TaskOfTResult_MethodAsync();
int intResult = await returnedTaskTResult;
// or, in a single statement
int intResult = await TaskOfTResult_MethodAsync();
They are actually different, why they use //or , in a single statement, just because it makes no different in their own example?
This is because when you are returning a running Task when you call Sleep() even when you're assigning to a variable.
The confusion is that the Task does not begin if you assign it to a variable (A, B, or C) until you call await A; but that's not true. As soon as you assign sleep(); to A, sleep() was called; therefore the Task in the sleep() method is running. Assigning it to a variable or not the Task begins when you call the method; because in the method you start the Task.
Knowing this; when you call:
await A;
await B;
await C;
A, B, and C, have already starting simultaneously... After awaiting A it is most likely B, and C have also completed or are milliseconds from completing.
There are situations where you can reference a Task that hasn't started yet but you would have to purposely return a non-running Task to do that.
To answer the edit to your question also.
Tasks have a method called GetAwaiter() which returns a TaskAwaiter. In C# when you write var task = sleep(); then you're assigning the actual Task to the task variable. All the same when you write await sleep(); the compiler does some cool stuff and it actually calls the Task.GetAwaiter() method; which is subscribed to. The Task will run and when it is complete the TaskAwaiter fires the continuation action. This can't be explained in a simple answer but to know the outer logic helps.
Among other things the TaskAwaiter implements ICriticalNotifyCompletion which in turn implements INotifyCompletion. Both have one method each, OnCompleted(Action) and UnsafeOnCompleted(Action) (you can guess which is which by naming convention).
Another thing to note is that Task.GetAwaiter() returns a TaskAwaiter but Task<TResult>.GetAwaiter() returns a TaskAwaiter<TResult>. There's not a strong difference in the two but there is a difference in the GetResult() method of the two tasks; which is what's called while marshalling back to the proper threading context. The TaskAwaiter.GetResult() returns void and the TaskAwaiter<TResult>.GetResult() returns TResult.
I feel like if I push further into this I'll have to write pages to explain it all in detail... Hopefully just explaining your question and pulling the curtain back a little bit will shed enough light to help you both understand and dig deeper if you're more curious.
Ok, so based on the comment below I want to describe my answer a little bit further.
I'll start this simple; let's just make a Task; one that isn't running, and look at it first.
public Task GetTask()
{
var task = new Task(() => { /*some work to be done*/ });
//Now we have a reference to a non-running task.
return task;
}
We can now call code like:
public async void DoWork()
{
await GetTask();
}
… but we'll be waiting forever; until the application ends, because the Task was never started. However; we could do something like this:
public async void DoWork()
{
var task = GetTask();
task.Start();
await task;
}
… and it will await the running Task and continue once the Task is complete.
Knowing this you can make as many calls to GetTask() as you like and you'll only be referencing Tasks that have not started.
In your code it's just the opposite, which is fine, as this is the most used way. I encourage you to make sure your method names notify the user of how you're returning the Task. If the Task is already running the most common convention is the end the method name with Async. Here's another example doing it with a running Task for clarity.
public Task DoTaskAsync()
{
var task = Task.Run(() => { /*some work to be done*/ });
//Now we have a reference to a task that's already running.
return task;
}
And now we will most likely call this method like:
public async void DoWork()
{
await DoTaskAsync();
}
However; note that if we simply want to reference the Task just like we did earlier, we can, the only difference is this Task is running where the one prior was not. So this code is valid.
public async void DoWork()
{
var task = DoTaskAsync();
await task;
}
The big take away is how C# handles the async / await keywords. async tells the compiler that the method is going to become a continuation of a Task. In short; the compiler knows to look for all await calls and put the rest of the method in a continuation.
The await keyword tells the compiler to call the Task.GetAwaiter() method on the Task ( and basically subscribe to the INotifyCompletion and ICriticalNotifyCompletion) to signal the continuation in the method.
And this I wanted to add just incase you weren't aware. If you do have more than one task that you want to await but would rather await one task as if they were all one then you can do that with Task.WhenAll() So instead of:
var taskA = DoTaskAsync();
var taskB = DoTaskAsync();
var taskC = DoTaskAsync();
await taskA;
await taskB;
await taskC;
You could write it a little cleaner like so:
var taskA = DoTaskAsync();
var taskB = DoTaskAsync();
var taskC = DoTaskAsync();
await Task.WhenAll(taskA, taskB, taskC);
And there are more ways of doing this sort of thing built in; just explore it.
Why the tasks are executed before Task.WhenAll??
If you see here, from the below code snippet, first Console.WriteLine("This should be written first.."); should be printed because I am awaiting the tasks beneath to it..
But if you see the output result, the Tasks method result is being printed before the above statement. Ideally, the tasks method should be executed when I await them, but it seems that- the tasks methods are executed the moment I add them in tasks list. Why is it so?
Would you please do let me know why is this happening??
Code:
public static async Task Test()
{
var tasks = new List<Task>();
tasks.Add(PrintNumber(1));
tasks.Add(PrintNumber(2));
tasks.Add(PrintNumber(3));
Console.WriteLine("This should be written first..");
// This should be printed last..
await Task.WhenAll(tasks);
}
public static async Task PrintNumber(int number)
{
await Task.FromResult(0);
Console.WriteLine(number);
}
Output
When you call an async method you get a "hot" task in return. That means that the task already started running (and maybe even completed) before you get to await them. That means that it's quite possible for the tasks to run and complete before the call to Task.WhenAll.
In your case however, while the PrintNumber is marked async it isn't asynchronous at all since you're using Task.FromResult. The synchronous part of an asynchronous method (which is the part until you await an asynchronous task) is always executed synchronously on the calling thread and is done before the call returns. When you use Task.FromResult you get a completed task so all your method is just the synchronous part and is completed before the call returns.
When you await a completed task (as is created by Task.FromResult, it completes synchronously. This means that in your example, nothing is actually happening asynchronously, which explains the order of execution.
If instead, you were to
await Task.Yield();
you'd see output more in line with your expectations.
Task.FromResult won't cause yield and the task will be executed on the same thread. To achieve what you want you can do this:
public static async Task Test()
{
var tasks = new List<Task>();
tasks.Add(PrintNumber(1));
tasks.Add(PrintNumber(2));
tasks.Add(PrintNumber(3));
Console.WriteLine("This should be written first..");
// This should be printed last..
await Task.WhenAll(tasks);
}
public static async Task PrintNumber(int number)
{
await Task.Yield();
Console.WriteLine(number);
}
If you want a Task or tasks to run after something else, its easiest to write your code accordingly.
public static async Task Test()
{
Console.WriteLine("This should be written first..");
// These should be printed last..
await Task.WhenAll(new[]
{
PrintNumber(1),
PrintNumber(2),
PrintNumber(3)
});
}
following on from your comment.
So we have some functions,
async Task<Customer> GetRawCustomer()
{
...
}
async Task<string> GetCity(Customer customer)
{
...
}
async Task<string> GetZipCode(Customer customer)
{
...
}
We could use them like this
var rawCustomer = await GetRawCustomer();
var populationWork = new List<Task>();
Task<string> getCity;
if (string.IsNullOrWhiteSpace(rawCustomer.City))
{
getCity = GetCity(rawCustomer);
populationWork.Add(getCity);
}
Task<string> getZipCode;
if (string.IsNullOrWhiteSpace(rawCustomer.City))
{
getZipCode = GetZipCode(rawCustomer);
populationWork.Add(getZipCode);
}
...
await Task.WhenAll(populationWork);
if (getCity != null)
rawCustomer.City = getCity.Result;
if (getZipCode != null)
rawCustomer.ZipCode = getZipCode.Result;
I was just experimenting to see what happens when a cold task (i.e. a Task which hasn't been started) is awaited. To my surprise the code just hung forever and "Finsihed" is never printed. I would expect that an exception is thrown.
public async Task Test1()
{
var task = new Task(() => Thread.Sleep(1000));
//task.Start();
await task;
}
void Main()
{
Test1().Wait();
Console.WriteLine("Finished");
}
Then I though perhaps the task can be started from another thread, so I changed the code to:
public async Task Test1()
{
var task = new Task(() => Thread.Sleep(1000));
//task.Start();
await task;
Console.WriteLine("Test1 Finished");
}
void Main()
{
var task1 = Test1();
Task.Run(() =>
{
Task.Delay(5000);
task1.Start();
});
task1.Wait();
Console.WriteLine("Finished");
}
But it is still blocked at task1.Wait(). Does anyone know if there is way to start a cold task after it has being awaited?
Otherwise it seems there is no point in being able to await a cold task, so perhaps the task should either be started when awaited or an exception should be thrown.
Update
I was awaiting the wrong task, i.e. the outer task returned by Test1 rather than the one newed inside it. The InvalidOperationException mentioned by #Jon Skeet was being thrown inside Task.Run however because the resulting task was not observed, the exception was not thrown on the main thread. Putting a try/catch inside Task.Run or calling Wait() or Result on the task returned by Task.Run threw the exception on the main console thread.
You're trying to start the task returned by the async method - that isn't the cold task that you started out with. Indeed, if you add some diagnostics to your Task.Run call, you'll see that an exception is thrown:
System.InvalidOperationException: Start may not be called on a promise-style task.
Here's an example showing what I think you were really trying to do:
using System;
using System.Threading;
using System.Threading.Tasks;
public class Test
{
static void Main(string[] args)
{
// Not using Task.Delay! That would be pointless
Task t1 = new Task(() => Thread.Sleep(1000));
Task t2 = Await(t1);
Console.WriteLine(t2.Status);
Console.WriteLine("Starting original task");
t1.Start();
Console.WriteLine(t2.Status);
t2.Wait();
Console.WriteLine(t2.Status);
}
static async Task Await(Task task)
{
Console.WriteLine("Beginning awaiting");
await task;
Console.WriteLine("Finished awaiting");
}
}
Note the use of Thread.Sleep instead of Task.Delay; unless you're using the result of Task.Delay, it basically does nothing. Using Thread.Sleep is emulating real work to be done in the other task.
As for why awaiting an unstarted task doesn't throw an exception - I think that's reasonable, to be honest. It allows for situations like the above to be valid, which may in some cases make life easier. (You may create a lot of tasks before starting them, for example - and you may want to start waiting for them to finish before you start them.)
Does anyone know if there is way to start a cold task after it has
being awaited?
You still can create a cold task from an async method and start it later, if that's what you want:
class Program
{
public static async Task Test1()
{
await Task.Delay(1000);
Console.WriteLine("Test1 is about to finish");
}
static void Main(string[] args)
{
var taskOuter = new Task<Task>(Test1);
var taskInner = taskOuter.Unwrap();
Task.Run(() =>
{
Thread.Sleep(2000);
// run synchronously
taskOuter.RunSynchronously();
// or schedule
// taskOuter.Start(TaskScheduler.Defaut);
});
taskInner.Wait();
Console.WriteLine("Enter to exit");
Console.ReadLine();
}
}