How does Task.Delay work exactly?

user_1939432 0
Viewed 14609

They say Task.Delay() is an async Thread.Sleep(). To test this I wrote below code. I expect to print immediately "One" then 3 seconds later result variable (15) will be printed. 2 seconds after this, "Two" will be printed. But it doesnt seem so. "One" is not immediately printed. "One" is printed 3 seconds later. Why does it wait 3 seconds to print "One" ? 

using System;
using System.Threading;
using System.Threading.Tasks;

namespace ConsoleApplication31
{
class Program
{
    public static int Multiply(int a, int b)
    {
        return a * b;
    }

    public static async Task DoRunAsync(int m, int n)
    {
        int result = await Task.Run(() => Multiply(m, n));
        await Task.Delay(3000);
        Console.WriteLine("One");
        Console.WriteLine(result);
    }

    static void Main(string[] args)
    {
        Task t = DoRunAsync(3, 5);
        Thread.Sleep(5000);
        Console.WriteLine("Two");
    }
}
}
asynchronous c#
3 Answers

The job of await is to suspend the current method until whatever awaitable you pass to it is complete. How, why and when that awaitable was created and started are of no relevance to what the await keyword accomplishes.

I think you're under the impression that await starts something, when in fact it's the opposite - it marks a point in your code where you wait for something to finish.

user_154980

It takes 3 seconds to print "One" because you await-ed Task.Delay too soon.

Change the code as follows to get the result that you expected:

int result = await Task.Run(() => Multiply(m, n));
var taskDelay = Task.Delay(3000); // No blocking here, so
Console.WriteLine("One");                              // printing can proceed.
await taskDelay; // This causes a block for the remainder of 3 seconds
Console.WriteLine(result);

When you start the delay task prior to printing "One" without await-ing it, subsequent WriteLine could complete without a delay.

user_3358580

Modifying your DoRunAsync method as follows will make things work as intended:

public static async Task DoRunAsync(int m, int n)
{
    int result = await Task.Run(() => Multiply(m, n));
    Console.WriteLine("One"); // before the delay, not after...
    await Task.Delay(3000);
    Console.WriteLine(result);
}

Your code behaves as follows:

await Task.Delay(3000); // Await 3 seconds...
Console.WriteLine("One"); // Once you are done awaiting, print the string...

If you await before printing the string, you can't expect it to be printed immediately...

user_7960850
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