我有一个看起来像这样的容器视图
<UserControl x:Class="Views.ContainerView">
<UserControl.Resources>
<ResourceDictionary>
<DataTemplate DataType="{x:Type viewmodels:AViewModel}">
<views:MyView />
</DataTemplate>
<DataTemplate DataType="{x:Type viewmodels:BViewModel}">
<views:MyView />
</DataTemplate>
<DataTemplate DataType="{x:Type viewmodels:CViewModel}">
<views:MyView />
</DataTemplate>
<DataTemplate DataType="{x:Type viewmodels:DViewModel}">
<views:MyView />
</DataTemplate>
</ResourceDictionary>
</UserControl.Resources>
<Grid>
<ListBox ItemsSource="{Binding Path=AvailableViewModels}"
SelectedItem="{Binding Path=CurrentViewModel}"
IsSynchronizedWithCurrentItem="True" />
<ContentControl Content="{Binding Path=CurrentViewModel}" />
</Grid>
</UserControl>
我所有的视图模型都继承了 BaseViewModel 所以我把我的视图变成了这个
<UserControl x:Class="Views.ContainerView">
<UserControl.Resources>
<ResourceDictionary>
<DataTemplate DataType="{x:Type viewmodels:BaseViewModel}">
<views:MyView />
</DataTemplate>
</ResourceDictionary>
</UserControl.Resources>
<StackPanel>
<ListBox ItemsSource="{Binding Path=AvailableViewModels}"
SelectedItem="{Binding Path=CurrentViewModel}"
IsSynchronizedWithCurrentItem="True" />
<ContentControl Content="{Binding Path=CurrentViewModel}" />
</StackPanel>
</UserControl>
认为它只会实例化一个 MyView 并在 ListBox.SelectedItem 更改时重新绑定视图模型。我是否正确理解了这种行为?这是首选做法吗?当我在视图之间切换时,如何验证我没有搅动内存?