通过功能<task>作为参数</task>

Question

I have a method:

public async Task Method(Func<Task> action)

This function basically checks if few seconds have passed since the last run time.

I tried:

Func<Task> timer = async () => { await Task.Delay(3000);};

and invoking it like that

Func<Task> check = await _className.Method(timer);

However it says

Cannot implicitly convert 'void' to 'System.Func<System.Threading.Tasks.Task>'

Basically I am not able to pass a delayed task to a function.

Answer 1

Here's a trivial example of calling Method with a Func<Task> .

Func<Task> timer = async () => { await Task.Delay(3000);};

await Method(timer);


async Task Method(Func<Task> action)
{
   await action();
}

Answer 2

I'm not sure exactly what you're trying to do, but this code works:

public class Program
{
    public static async Task Main(string[] args)
    {
        var sw = Stopwatch.StartNew();
        Func<Task> timer = async () => { await Task.Delay(3000); };
        var theTask = Method(timer);
        await theTask;
        Console.WriteLine(sw.ElapsedMilliseconds); // ~3000
    }

    public static async Task Method(Func<Task> action)
    {
        await action();
    }
}

I just put in var theTask = Method(timer); to illustrate that you can assign the return value to a Task before awaiting it.

Answer 3

Okay, so it works like tymtam already said - it looks like my problem was trying to assign end result to Func, i guess i need more sleep.

Thank you for the help tho

EDIT I needed this:

Func<Task> check = async () => await _className.Method(timer);