WPF Unfreeze UI when showing heavy window - wpf

Question

I have heavy window (Devexpress UI), and, when I call window.Show() in my ICommand body, UI freezes. I want to show some progress while window loaded.
I was trying to call it in async method, but could not find any way to call Show in async way:
I was try Task.Run but it fail with exception of different UI threads.
I was try to call Application.Dispatcher.BeginInvoke but this also freezes UI.
Little adjustment:
I was wrong about Show() freezes. UI freeze at constructor call: var window = new ProductsPricesWindow()

Answer 1

If your execute some time-consuming work in another thread, then you must synchronize a result of another thread with UI thread. To synchronize two threads(new thread and UI thread), it is necessary to use Dispatcher.
As MSDN says:
Only one thread can modify the UI thread. But how do background
threads interact with the user? A background thread can ask the UI
thread to perform an operation on its behalf. It does this by
registering a work item with the Dispatcher of the UI thread. The
Dispatcher class provides two methods for registering work items:
Invoke and BeginInvoke. Both methods schedule a delegate for
execution. Invoke is a synchronous call – that is, it doesn’t return
until the UI thread actually finishes executing the delegate.
BeginInvoke is asynchronous and returns immediately.
For example:
xaml:
<TextBox Name="textBox"/>
in code-behind:
Task.Run(()=> {
Thread.Sleep(5000);//imitate time consuming work
Dispatcher.BeginInvoke(System.Windows.Threading.DispatcherPriority.Render,
new Action(() => {
textbox.Text="Hello World!:)"}));
});
In the above example, we create a new thread (Task.Run(...)) and synchronize result of new thread with UI thread by calling Dispatcher(Dispatcher.BeginInvoke(...))
UPDATE:
You should call new window on UI thread. So the code would be:
var yourWindow=new YourWindow();
yourWindow.ShowDialog();
However, as you did not show any code located in constructor of YourWindow, I try to suppose that there is some method which is time consuming. Then your code should look like this:
public class YourWindow()
{
public YourWindow()
{
Task.Run(()=> {
TimeConsumingMethod();
});
}
private void TimeConsumingMethod()
{
Thread.Sleep(5000);//imitate time consuming work
Dispatcher.BeginInvoke(System.Windows.Threading.DispatcherPriority.Render,
new Action(() => {
textbox.Text="Hello World!:)"}
}
}

Answer 2

Looks like this is not possible to make UI responsible when other UI component initialized.
So in my case solution is just allow UI to show in BusyIndicator (I use Xceed.Wpf.Toolkit) current state.
To do this I just put await Task.Delay(100); before creating window.
Also was helpful to pre-initialize creating window in background thread that starts on my app start. This makes creatng new window more responsible and even makes BusyIndicator live sometime.
public void Init()
{
var thr = new Thread(
async () =>
{
BusyContent = "Loading...";
try
{
// pre-init dx
await Application.Current.Invoke(
() =>
{
var window = new Views.ProductsPricesWindow();
window.Close();
}
);
...
}
finally
{
NotBusy();
}
}
);
thr.Name = nameof(Init);
thr.Start();
}
public async Task AddProductsPrices(Window p)
{
try
{
BusyContent = "Loading...";
await Task.Delay(100);
var window = new ProductsPricesWindow();
window.Show();
}
finally
{
NotBusy();
}
}
Application.Current.Invoke is extender:
public async static Task Invoke(this Application app, Action action, DispatcherPriority priority = DispatcherPriority.Normal)
{
if (app != null)
{
await app.Dispatcher.BeginInvoke(
priority,
action
);
}
else
action();
}
XAML
<extToolkit:BusyIndicator
IsBusy="{Binding IsBusy, Mode=OneWay}"
>
<extToolkit:BusyIndicator.BusyContent>
<TextBlock
Text="{Binding BusyContent}"
FontSize="72"
/>
<Grid>
...
</Grid>
</extToolkit:BusyIndicator.BusyContent>
</extToolkit:BusyIndicator>