.NET: Async: return value

Question

I have 2 async functions that I execute and return an integer. When I try to access the return value, the IDE complains that the tasks are void type, not int type.

private async Task<int> func01(int _i1, int _i2)
{
    return await Task.Run(async () => {                
            await Task.Delay(1000);
            return 1;
        });

}

private async Task<int> func02(string _i1, string_i2)
{
    return await Task.Run(async () => {                
            await Task.Delay(3000);
            return 1;
        });

}

Task _t1 = func01(1, 3);
Task _t2 = func02("Hello", "World");
int _r1 = await _t1; //here, IDE says that _t1 and _t2 are void types
int _r2 = await _t2;

How do I return a value from an async task?

You are upcasting the return values of the functions to Task instead of Task<int> — UnholySheep
– UnholySheep, Commented Apr 21, 2022 at 20:08

Onurkan Bakırcı · Accepted Answer · 2022-04-21 20:29:31Z

6

Because return type of functions are not Task, its return Task<int>. So you have to use;

Task<int> _t1 = func01(1, 3);
Task<int> _t2 = func02("Hello", "World");

Tip: You can also use var keyword. I think its better.

var _t1 = func01(1, 3);
var _t2 = func02("Hello", "World");

edited Apr 21, 2022 at 20:29

answered Apr 21, 2022 at 20:09

Onurkan Bakırcı

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2 Comments

Nigel Over a year ago

var will work too. I personally prefer declaring all local variables as var

Onurkan Bakırcı Over a year ago

I update answer.

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