External initialization of ViewModel (code cleanup)

Author: apdevelop

Src/ScreenGrid/App.xaml.cs

@@ -16,7 +16,10 @@ protected override void OnStartup(StartupEventArgs e)
             base.OnStartup(e);
             this.DispatcherUnhandledException += AppDispatcherUnhandledException;
 
-            (new ScreenGrid.Views.ScreenGridWindow()).Show();
+            (new Views.ScreenGridWindow
+            {
+                DataContext = new ViewModels.ScreenGridViewModel()
+            }).Show();
         }
 
         private void CurrentDomainUnhandledException(object sender, UnhandledExceptionEventArgs e)

Src/ScreenGrid/Views/ScreenGridWindow.xaml.cs

@@ -21,16 +21,11 @@ public ScreenGridWindow()
             // Making text sharp
             TextOptions.SetTextFormattingMode(this, TextFormattingMode.Display);
 
-            // TODO: external ViewModel
             this.InitializeComponent();
-            this.vm = new ViewModels.ScreenGridViewModel();
-            this.DataContext = this.vm;
         }
 
         private ContextMenu mainMenu;
 
-        private ViewModels.ScreenGridViewModel vm;
-
         private void Window_Loaded(object sender, RoutedEventArgs e)
         {
             this.CreateContextMenu();
@@ -75,7 +70,7 @@ private void CreateContextMenu()
                     }
 
                     menuItem.IsChecked = true;
-                    this.vm.GridMode = (Models.Grids.GridType)((s as MenuItem).Tag);
+                    (this.DataContext as ViewModels.ScreenGridViewModel).GridMode = (Models.Grids.GridType)((s as MenuItem).Tag);
                 };
 
                 this.mainMenu.Items.Add(menuItem);