Why doesn't this example work with async await?

Question

.NET

class Program
{
    static void Main(string[] args)
    {
        Task<int> t = Run();
        Console.WriteLine("Main");
        Console.ReadLine();
    }

    static async Task<int> Run()
    {
        Console.WriteLine("Hello from begining \"Run\" method");
        int num = 1000000000;
        long res = await new TaskFactory().StartNew(() => { return SumZeroToNum(num); });
        Console.WriteLine($"Result: {res}");
    }

    static long SumZeroToNum(int num)
    {
        long result = 0;
        for (int i = 0; i < num; i++)
        {
            result += i;
        }
        return result;
    }
}

When trying to compile this code, the studio throws error CS0161 for the Run method : not all code paths return a value.
In my understanding, after creating a task, control is returned to the Main method, i.e. returns Task. Then why does the error occur?

Answer 1

Fix the Run method

static async Task Run()
    {
        Console.WriteLine("Hello from begining \"Run\" method");
        int num = 1000000000;
        long res = await new TaskFactory().StartNew(() => { return SumZeroToNum(num); });
        Console.WriteLine($"Result: {res}");
    }

and the line to call this method:
Task t = Run();

Answer 2

You have several mistakes here.
First, you Run()indicated in the method signature that it returns an object of the type, Task,but you didn't return it. If you want it to return a task, then you need to create it

var task = new Task(() => async
{ 
  Console.WriteLine("Hello from begining \"Run\" method");
    int num = 1000000000;
    long res = await new TaskFactory().StartNew(() => { return SumZeroToNum(num); });
    Console.WriteLine($"Result: {res}");
});

and then return
But having done in the Main Run () method, you will only receive the task object, it will not be executed automatically for you. To do this, you will need to do t.Start();
If you want to call the method through async, then inside you need to do this

return Task.Run(() =>
{
    ....
})

Then it will be possible to run it asynchronously
Or synchronously
Run().Result;