Logo Questions Linux Laravel Mysql Ubuntu Git Menu
 

Cannot convert lambda expression to type "..." because it is not a delegate type

Good day! I am trying to write an anonymous method using lambda expressions which would return an object from an async task. I would like to do this in the constructor, so that is the reason I can't make its parent method async.

The ReadJsonAsync method returns a Session object. I will show you the relevant code:

Session session;
fileService = new FileService();
session = async () =>  { return await fileService.ReadJsonAsync() };

Thanks in advance!

like image 932
masm64 Avatar asked Nov 15 '14 14:11

masm64


2 Answers

If you want an Anonymous Method, you'll have to declare one which returns a Task<Session> as it is marked with the async modifier, hence must return a void (only for async event handlers), Task or Task<T> :

Func<Task<Session>> anonFunction = async () => await fileService.ReadJsonAsync();

If all you do is run ReadJsonAsync, you may also save yourself the state machine generation like so:

Func<Task<Session>> anonFunction = fileService.ReadJsonAsync;

Then you can await on it higher up the call stack:

Func<Task<Session>> anonFunction = fileService.ReadJsonAsync;
await anonFunction();
like image 151
Yuval Itzchakov Avatar answered Nov 14 '22 23:11

Yuval Itzchakov


To add to Yuval's useful answer, if you just want to await an inline function, then the magic syntax is:

await ((Func<Task>)(async () =>
{
    //async function code
}
))();

Note the extra brackets on the end to call the lambda immediately after declaration. Obviously if your function returns a type, then it would be Func<Task<Whatever>>

Useful if you're using Task.WhenAny() for example to await both an inline function and a timeout task.

like image 30
GazTheDestroyer Avatar answered Nov 14 '22 23:11

GazTheDestroyer