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Home/ Questions/Q 6139479
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T17:59:41+00:00 2026-05-23T17:59:41+00:00

This code is from my C# WPF application: public MainWindow() { InitializeComponent(); status(Getting dl

  • 0

This code is from my C# WPF application:

public MainWindow()
{
    InitializeComponent();
    status("Getting dl links");
    getLinks();
}

The procedure getLinks currently displays a couple of links in a messagebox. Those links are displayed in a messagebox before the WPF application becomes visible.
In this is case not a problem, but how would I show progress (like a progressbar) of any
procedure I want to load at startup?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-23T17:59:42+00:00Added an answer on May 23, 2026 at 5:59 pm

    Here is an example on how you can do it. To simplify it a bit, I added the controls directly in the MainWindow constructor, but I would prefer to do this with XAML.

    public MainWindow()
    {
        InitializeComponent();
    
        var progressBar = new ProgressBar();
        progressBar.Height = 40;
        progressBar.Width = 200;
        progressBar.Margin = new Thickness(100, 100, 100, 100);
    
        Task.Factory.StartNew(() =>
        {
            // getLinks();
            for (var i = 0; i < 5; i++)
            {
                Dispatcher.Invoke(new Action(() => { progressBar.Value += 20; }));
    
                Thread.Sleep(500);
            }
        });
    
        var stackPanel = new StackPanel();
        stackPanel.Children.Add(progressBar);
    
        Content = stackPanel;
    
    }
    

    I first add a ProgressBar somewhere on the interface to make it visible for this demo and then I add it to a new StackPanel, it could be any panel at all, in this case it doesn’t matter.

    To load the links on another thread, I create a new Task, this is a part of the TPL (Task Parallel Library) in .NET 4.0. In this case I am simulating that getLinks() takes 5 * 500 milliseconds to run and that it in fact is five links that will be loaded, hence 20% each iteration.

    What I do then is that I add 20 to the progressBar value, which indicates that it increased with 20%.

    This line might confuse you a bit

    Dispatcher.Invoke(new Action(() => { progressBar.Value += 20; }));

    But it is in fact quite common when you do cross-thread programming with GUI. So the problem is that you are on another thread here, we started of a Task that will run on a separate thread, and you cannot update your UI thread from another thread. So what you need is something called a Dispatcher, and this is accessable from within your Window-class.

    Then you Invoke an action on it, which means that you simply say “Run this piece of code on this thread for me”.

    And if you want to display a MessageBox when everything is loaded, you can simply add a MessageBox.Show("Loaded!"); after the for-loop.

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