ViewModel instatiated twice with MEF

Question

I'm trying to create a simple modular MVVM application with MEF. I have a ViewModel class and a UserControl as the View. I connect the two through a DataTemplate, like this:

<DataTemplate DataType="{x:Type local:MyViewModel}"> <local:MyView /> </DataTemplate>

In the View, I define the ViewModel as a StaticResource, to make binding simple:

<UserControl.Resources> <local:MyViewModel x:Key="ViewModel" /> </UserControl.Resources>

Then I bind like this:

<Grid DataContext="{StaticResource ResourceKey=ViewModel}"> <TextBlock Text="{Binding Text}" /> </Grid>

This all works as intended without MEF. However, as I am aiming for modularity, I use MEF to discover my ViewModel classes. I have an Export attribute on my ViewModel class:

[Export(typeof(MyViewModel))] public class MyViewModel { // ... }

and I use MEF to dynamically load the ViewModel into my shell in App.xaml.cs:

private void Application_Startup(object sender, StartupEventArgs e) { var shell = new MainWindow(); var catalog = new AssemblyCatalog(this.GetType().Assembly); var container = new CompositionContainer(catalog); shell.Contents.ViewModel = container.GetExportedValues<MyViewModel>().First(); shell.Show(); }

Now, at this point, MEF creates an instance of my ViewModel when it loads the vm, and my View creates another instance when it declares the vm as a resource. (This is easily checked by setting a breakpoint in the constructor.)

The question is, how should I pass the instance created by MEF to my resource declaration? Can I declare that specific instance as resource?

DropBox link with full code: https://www.dropbox.com/sh/pbdl029d26sx7gl/AAA6po50dLjbJSoNPBhCyWZ3a?dl=0

Modified my infrastructure using DataContextSpy based on this answer: stackoverflow.com/a/5402653/5219911 Leaving the question open for a bit, in case someone comes up with another soulition, but this seems to work at the moment. — Attila Bertok
– Attila Bertok, Commented Sep 29, 2015 at 17:49

cscmh99 · Accepted Answer · 2015-09-28 16:59:27Z

The creation of the MyViewModel is purely based on the sequence of how your program execution but you can set the CreationPolicy to make your instance become a singleton so that both your code and the resources both referring to the same instance.

[Export(typeof(MyViewModel)), PartCreationPolicy(CreationPolicy.Shared)]

Side note: To use MEF, Microsoft hide the implementation of CompositionInitializer and CompositionHost from .Net Framework for some reason. Try to google up and import the 2 classes from Microsoft instead of using the CompositionContainer directly. You'll have better experience in using MEF with those.

This unfortunately only tells MEF to instantiate my viewmodel only once.The issue is that both MEF and the WPF resource lookup tries to instantiate said viewmodel. If I created the view directly, I'd be able to add a viewmodel resource. However, as the view is created from a DataTemplate, I don't have a reference to the view to directly acces its resources.

Community · Accepted Answer · 2017-05-23 12:14:31Z

Okay, so, what I had to do was this.

I had two instances because once MEF instantiated the ViewModels when it imported them, and then another time WPF created them, when it created the ViewModel resource. I figured that the solution would be not directly creating the resource, but had no idea how I could manage to do that. Then along came Resource Injection and then DataContextSpy, from this question here: https://stackoverflow.com/a/5402653/5219911

And here's the direct link to the topic: http://www.codeproject.com/Articles/27432/Artificial-Inheritance-Contexts-in-WPF

I am now using a resource that is a DataContextSpy, through which I can reach out to the ViewModel instance that is used when creating the DataTemplate.

In my View's resources, I define and then I just set the root element's DataContext to this resource: DataContext="{Binding Source={StaticResource ResourceKey=ViewModel}, Path=DataContext}"

Now, unfortunately, I don't get Intellisense support with this alone, as DataContextSpy is sort of a proxy to the real DataContext, so I do have to manually set up the design time DataContext type by using: d:DataContext="{d:DesignInstance Type=viewModel:MyViewModel}"

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