Answers

What is the output of the program below? Give an explanation.

Naimish Makwana

3y4.4kC#

 public class TestAsync
{
    private static string result;
    static void Main()
    {
        PrintAsync();
        Console.WriteLine(result);
    }
    static async Task<string> PrintAsync()
    {
        await Task.Delay(5);
        result = "Hello world!";
        return "This is example";
    }
}

5 Answers

This program will not return any result because of async method call without using of await.

Rajeev Kumar3y

In the Main() method, PrintAsync() is called, but since it is an asynchronous method, it returns immediately before its task is complete. The Console.WriteLine(result) statement executes and prints the current value of result, which is still null because the task returned by PrintAsync() hasn’t finished yet.

When the task finally completes after the 5 millisecond delay, it sets the value of result to “Hello world!”, but this happens after the Console.WriteLine(result) statement has already executed. Therefore, the output of the program is null.

Tuhin Paul3y

The output of this program will be null

Tuhin Paul3y

This will not show any result on the screen.
I agree with Jignesh

Just to add we can also add the async and Task keyword into the Main method so we can add the asyncbefore the PrintAsync method. Although this is supported starting at C# 7.1.

static async Task Main(string[] args)
{
    var r = await PrintAsync();
    Console.WriteLine(r); //This is example
    Console.WriteLine(result); //Hello World
    Console.ReadLine();
}

Hope this helps. Thanks.

Jin Necesario3y

Above code snippet will not return any result because of async method call without await.

if you are expeting to execute then you can call in two ways,

var r = PrintAsync().Result;

Another way you have to call with await keyword then will return result as “Hello world!”

Jignesh Kumar3y