The reason is your result assigned to the returning Task which represents continuation of your method, and you have a different Task in your method which is running, if you directly assign Task like this you will get your expected results:
var task = Task.Run(() =>
{
for (int i = 10; i < 432543543; i++)
{
// just for a long job
double d3 = Math.Sqrt((Math.Pow(i, 5) - Math.Pow(i, 2)) / Math.Sin(i * 8));
}
return "Foo Completed.";
});
while (task.Status != TaskStatus.RanToCompletion)
{
Console.WriteLine("Thread ID: {0}, Status: {1}", Thread.CurrentThread.ManagedThreadId,task.Status);
}
Console.WriteLine("Result: {0}", task.Result);
Console.WriteLine("Finished.");
Console.ReadKey(true);
The output:
Consider this for better explanation: You have a Foo method,let's say it Task A, and you have a Task in it,let's say it Task B, Now the running task, is Task B, your Task A awaiting for Task B result.And you assing your result variable to your returning Task which is Task A, because Task B doesn't return a Task, it returns a string. Consider this:
If you define your result like this:
Task result = Foo(5);
You won't get any error.But if you define it like this:
string result = Foo(5);
You will get:
Cannot implicitly convert type 'System.Threading.Tasks.Task' to 'string'
But if you add an await keyword:
string result = await Foo(5);
Again you won't get any error.Because it will wait the result (string) and assign it to your result variable.So for the last thing consider this, if you add two task into your Foo Method:
private static async Task<string> Foo(int seconds)
{
await Task.Run(() =>
{
for (int i = 0; i < seconds; i++)
{
Console.WriteLine("Thread ID: {0}, second {1}.", Thread.CurrentThread.ManagedThreadId, i);
Task.Delay(TimeSpan.FromSeconds(1)).Wait();
}
// in here don't return anything
});
return await Task.Run(() =>
{
for (int i = 0; i < seconds; i++)
{
Console.WriteLine("Thread ID: {0}, second {1}.", Thread.CurrentThread.ManagedThreadId, i);
Task.Delay(TimeSpan.FromSeconds(1)).Wait();
}
return "Foo Completed.";
});
}
And if you run the application, you will get the same results.(WaitingForActivation) Because now, your Task A is waiting those two tasks.