Visual studio bug when building - wpf

Question

I have this wierd bug. When I build my project, it always pops up two instances of my singleton window. Here is video of it happening. Though I don't think it will help anyting, here is most of xaml of the window. In code behind are just events for mouse movement and a closing event.
<Window x:Class="Scoreboard.OutputWindow"
d:DataContext="{d:DesignInstance viewModel:ControlPanel}"
Title="OutputWindow" Height="375" Width="540"
Background="{Binding AppData.Settings.Color, Converter={converters:ColorToBrushConverter}}"
MouseLeftButtonDown="football_MouseLeftButtonDown"
MouseLeftButtonUp="football_MouseLeftButtonUp"
MouseMove="football_MouseMove">
<Window.InputBindings>
<KeyBinding Modifiers="Control" Key="F" Command="{Binding FullScreenCommand}" />
</Window.InputBindings>
<Grid ...>
</Grid>
</Window>
This window is a singleton in my ViewModel. I access it just from there. It happens on both of my two PCs.
EDIT:
When I remove it from the ViewModel, it doesn't happen anymore. But I sort of need it there.
EDIT2: Here is new project with that window. It doesn't behave exactly the same, but when you do some changes so it builds on debug, then on sturtup it creates another OutputWindow which isn't caught by breakpoint in constructor.

Answer 1

The root cause of your problem is the following line in your ViewModel:
public ViewModel()
{
OutputWindow.IsEnabled = true;
}
This implicitly creates a new OutputWindow() in the getter of the property:
private OutputWindow _outputWindow;
public OutputWindow OutputWindow =>
_outputWindow ?? (_outputWindow = new OutputWindow { DataContext = this });
The constructor then looks like this:
public OutputWindow()
{
InitializeComponent();
Show();
}
This wouldn't be so bad at the moment but every time your MainWindow is instantiated, a new instance of ViewModel is implicitly created as well because of this code in the MainWindow:
<Window.DataContext>
<local:ViewModel />
</Window.DataContext>
Okay, that explains why the OutputWindow is showing up on start - but why does it show on build?
(Or in my case even if you open the XAML code of MainWindow.xaml in Visual Studio.)
If found the culprit and it's this little gem here which originates from devenv.exe:
TestApplication.exe!TestApplication.OutputWindow.OutputWindow() Line 15
TestApplication.exe!TestApplication.ViewModel.OutputWindow.get() Line 6
TestApplication.exe!TestApplication.ViewModel.ViewModel() Line 10
...
mscorlib.dll!System.Activator.CreateInstance(System.Type type, bool nonPublic)
...
Microsoft.VisualStudio.DesignTools.Platform.dll!Microsoft.VisualStudio.DesignTools.Platform.InstanceBuilders.ClrObjectInstanceBuilder.Instantiate(Microsoft.VisualStudio.DesignTools.Platform.InstanceBuilders.IInstanceBuilderContext context, Microsoft.VisualStudio.DesignTools.Platform.InstanceBuilders.ViewNode viewNode = {Microsoft.VisualStudio.DesignTools.Platform.InstanceBuilders.ViewNode}) Unknown
Microsoft.VisualStudio.DesignTools.Platform.dll!Microsoft.VisualStudio.DesignTools.Platform.InstanceBuilders.ViewNodeManager.CreateInstance(Microsoft.VisualStudio.DesignTools.Platform.InstanceBuilders.IInstanceBuilder builder, Microsoft.VisualStudio.DesignTools.Platform.InstanceBuilders.ViewNode viewNode) Unknown
Microsoft.VisualStudio.DesignTools.Platform.dll!Microsoft.VisualStudio.DesignTools.Platform.InstanceBuilders.ViewNodeManager.Instantiate(Microsoft.VisualStudio.DesignTools.Platform.InstanceBuilders.ViewNode viewNode = {Microsoft.VisualStudio.DesignTools.Platform.InstanceBuilders.ViewNode}) Unknown
...
XDesProc.exe!Microsoft.VisualStudio.DesignTools.DesignerContract.Isolation.DesignerProcess.RunApplication()
...
mscorlib.dll!System.Threading.ThreadHelper.ThreadStart()
So, why does it happen?
On build or on open of the XAML designer, Visual Studio starts a thread and instructs XDesProc.exe to create a new instance of your MainWindow.
(XDesProc.exe is the component that is responsible for the XAML designer in Visual Studio and perhaps the designer should refresh if you build a project.)
And from there on, the way is as shown above:
A new ViewModel is created in the implicit DataContext of the MainWindow which implicitly creates a new instance of the OutputWindow in the getter of the property which in turn calls Show() on itself. And voilà, there you have your OutputWindow on screen!
In some cases there are even two OutputWindows showing but that's only because the designer (or maybe another component) is called two times.
So, bottom line - what do you need to fix your code?
There are multiple possible solutions.
Most important you should stop the implicit creation of the OutputWindow in the ViewModel. You can achieve this by doing one (or more) of the following:
Remove OutputWindow.IsEnabled = true; in the constructor of ViewModel.
Don't call Show(); in the constructor of OutputWindow.
Don't instantiate your ViewModel implicitly using <Window.DataContext><local:ViewModel />.
I think it's fair to say that it's just plain wrong to implicitly show a window if you just create an instance of a simple ViewModel. So, if you break this unlucky chain of implicit events then your problem will be gone forever.