Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
443 views
in Technique[技术] by (71.8m points)

c# - Async wait block Main UI

I am using the new async await features to upgrade from backgroundworker in C#. In the following code I am trying to replicate the execution of multiple tasks with ContinueWith method.

        Task t1 = new Task
        (
            () =>
            {
                Thread.Sleep(10000);

                // make the Task throw an exception
                MessageBox.Show("This is T1");
            }
        );

        Task t2 = t1.ContinueWith
        (
            (predecessorTask) =>
            {

                if (predecessorTask.Exception != null)
                {
                    MessageBox.Show("Predecessor Exception within Continuation");

                    return;
                }

                Thread.Sleep(1000);
                MessageBox.Show("This is Continuation");
            },

            TaskContinuationOptions.AttachedToParent | TaskContinuationOptions.OnlyOnRanToCompletion
        );

        t1.Start();

        try
        {
           t1.Wait(); <------ Comment 
           t2.Wait(); <------ Comment 
        }
        catch (AggregateException ex)
        {
            MessageBox.Show(ex.InnerException.Message);
        }  

My question is when I comment t1.wait and t2.wait Tasks are not blocking UI. However when I uncomment t1.wait and t2.wait UI blocks until thread is completed. The desired behavior is to catch errors in try/catch block without blocking UI. What Am I missing?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

If this is running within a UI event handler, you can add the async modifer to the method signature and change t1.Wait() to await t1. This will return control to the UI thread, and when the Thread.Sleep has completed, the continuation will execute and any exceptions will be caught.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...