What must I do to make my methods awaitable?

Question

How can I roll my own async awaitable methods?

I see that writing an async method is easy as pie in some cases:

private async Task TestGeo()
{
    Geolocator geo = new Geolocator();
    Geoposition pos = await geo.GetGeopositionAsync();
    double dLat = pos.Coordinate.Latitude;
    double dLong = pos.Coordinate.Latitude;
}

...but sadly also see that not just any method can be made async willy-nilly, though; to wit: this doesn't work:

private async Task TestAsyncAwait()
{
    int i = await TaSLs_Classes.TASLsUtils.GetZoomSettingForDistance(5);
}

...it stops me with the compile error, "Cannot await int"; a hint at design time similarly tells me, "type 'int' is not awaitable"

I also tried this with the same results:

    Task<int> i = await TaSLs_Classes.TASLsUtils.GetZoomSettingForDistance(5);

What must I do to make my methods awaitable?

UPDATE

As Linebacker and S. Cleary indicated (any relation to that cat who used to be on KNBR?), this works:

int i = await Task.Run(() => TaSLs_Classes.TASLsUtils.GetZoomSettingForDistance(5));

...that is to say, it compiles -- but it never "moves."

At runtime, it tells me I should "await" the CALL to TestAsyncAwait(), but if I do that, it doesn't compile at all...

Solution 1

You only need to return an awaitable. Task/Task<TResult> is a common choice; Tasks can be created using Task.Run (to execute code on a background thread) or TaskCompletionSource<T> (to wrap an asynchronous event).

Read the Task-Based Asynchronous Pattern for more information.

Solution 2

Your method

private async Task TestAsyncAwait()
{
    int i = await TaSLs_Classes.TASLsUtils.GetZoomSettingForDistance(5);
}

should be written like this

private async Task TestAsyncAwait()
{
    Task<int> t = new Task<int>(() =>
    {
        return TaSLs_Classes.TASLsUtils.GetZoomSettingForDistance(5);
    });
    t.Start();
    await t;
}

If you need to return the int, replace the Task type:

private async Task<int> TestAsyncAwait()
{
    Task<int> t = new Task<int>(() =>
    {
        return TaSLs_Classes.TASLsUtils.GetZoomSettingForDistance(5);
    });
    t.Start();
    return await t;
}

Solution 3

var something = Task<int>.Factory.StartNew(() => 0);
something.Wait();
int number = something.Result;

Solution 4

You need to either consume the awaited return or return Task<Type of await call>

1. private async Task<Geoposition> TestGeo()
   {
       Geolocator geo = new Geolocator();
       return await geo.GetGeopositionAsync();
   }
   

2. private async Task<int> TestAsyncAwait()
   {
       return await TaSLs_Classes.TASLsUtils.GetZoomSettingForDistance(5);
   }
   

View more solutions

Comments

[L.B]

int i = await Task.Run(() => TaSLs_Classes.TASLsUtils.GetZoomSettingForDistance(5));

[Christian]

I recommend you read TAP

[B. Clay Shannon-B. Crow Raven]

So Lambdas are not just de rigeur, but obligatory now? I've viewed them, at least up to now, as clever-clever tricks.

[Stephen Cleary]

Are you sure you want to make this awaitable?

[B. Clay Shannon-B. Crow Raven]

No, I'm just experimenting - I want to know how to make methods awaitable so that I can "do it until the cows come home" or "without even thinking (much) about it" when I need to.

[Jeson Martajaya]

The question is about #2, and for #2, the error says "Type 'int' is not awaitable"

[aaronburro]

Downvoted because the OP is clearly trying to do this asynchronously

[philologon]

Thanks to aaronburro for adding a downvote explanation. This is too rare.

[Denis G. Labrecque]

The link is dead

[Marcelo Scofano Diniz]

Rare and useful.