WPF update UI from async method

Question

In public MainWindow() I call TestTask3(10000);. A TextBlock Text is bound to Answer in the XAML. The get is called and the initial value is displayed. I see the set call NotifyPropertyChanged. But the get is not called a second time to get the new value. I also have a button that changes Answer and that does change the TextBlock.

How to fix this or another approach? I want to keep the UI reactive and update some UI elements after a delay.

private int answser = -2;
public int Answer
{
    get { return answser; }
    set
    {
        if (answser != value)
        {
            answser = value;
            NotifyPropertyChanged("Answer");
        }
    }
}
public async void TestTask3(int delay)
{
    Debug.WriteLine($"TestTask3");
    int answer = -1;
    int i = await Task.Run(() =>
    {
        // … do compute-bound work here  
        Task.Delay(delay);
        answer = -1;
        return answer;
    });
    Debug.WriteLine($"TestTask3   {i}");
    Answer = answer;
    //return answer;
}

You shouldn’t use the same names for local variables and class fields, it’s very confusing. Have you tried putting a breakpoint in the setter and stepping through? — Dave M
– Dave M, Commented May 11, 2018 at 1:37
Yeah this is confusing, i cant see how on earth the property Answer would get updated twice anyway (if i'm understanding you correctly) — TheGeneral
– TheGeneral, Commented May 11, 2018 at 1:37
@DaveM As stated I see NotifyPropertyChanged called. That is how is see that. — paparazzo
– paparazzo, Commented May 11, 2018 at 1:41
@TheGeneral It gets updated if I change Answer in a button click event. — paparazzo
– paparazzo, Commented May 11, 2018 at 1:42

Clemens · Accepted Answer · 2018-05-11 05:47:16Z

4

Task.Delay(delay) is awaitable method.

Following solution works for me. I took a long loop calculation and it returns me result on UI after 10 seconds. Tested and working.

public async void TestTask3(int delay)
{
    int answer = -1;
    int i = await Task.Run(async () =>
    {
        // … do compute-bound work here  
        for (int j = 0; j < 100000000; j++)
        {
            answer += j;
        }

        await Task.Delay(delay);

        //answer = -1;
        return answer;
    });
    Answer = answer;
    //return answer;
}

Usage:

private async void Button_Click(object sender, RoutedEventArgs e)
{
    await TestTask3(10000);
}

edited May 11, 2018 at 5:47

Clemens

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answered May 11, 2018 at 1:42

Gaurang Dave

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6 Comments

paparazzo Over a year ago

So it was missing an async. Not sure how that broke it but I am just glad it is fixed.

Gaurang Dave Over a year ago

I am glad to help. Cheers :)

Clemens Over a year ago

@paparazzo Since you should usually avoid async void methods (except event handlers), you may declare public async Task<int> TestTask3(int delay) and make it return answer. Then call it like Answer = await TestTask3(10000);

paparazzo Over a year ago

@Clemens That syntax throws a design time error for me. ` int ans = TestTask3(1000, token).Result;` is valid syntax but is hangs on that statement.

Clemens Over a year ago

It must be int ans = await TestTask3(...). Don't use Result. See here: stackoverflow.com/q/27464287/1136211

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