Using Bindingoperations.Enablecollectionsynchronization

Using BindingOperations.EnableCollectionSynchronization

All the examples I've seen on Stack Overflow for this get it wrong. You must lock the collection when modifying it from another thread.

On dispatcher (UI) thread:

_itemsLock = new object();
Items = new ObservableCollection<Item>();
BindingOperations.EnableCollectionSynchronization(Items, _itemsLock);

Then from another thread:

lock (_itemsLock)
{
    // Once locked, you can manipulate the collection safely from another thread
    Items.Add(new Item());
    Items.RemoveAt(0);
}

More information in this article: http://10rem.net/blog/2012/01/20/wpf-45-cross-thread-collection-synchronization-redux

How does EnableCollectionSynchronization work?

You should synchronize all access to the collection using the same lock, i.e. you should lock around the call to Add:

lock (itemsLock)
    items.Add("foo");

The documentation is pretty clear on this:

To use a collection on multiple threads, one of which is the UI thread that owns the ItemsControl, an application has the following responsibilities:

• Choose a synchronization mechanism.
• Synchronize all access from the application to the collection using that mechanism.
• Call EnableCollectionSynchronization to inform WPF of the mechanism.
• ...

How does BindingOperations.EnableCollectionSynchronization work, and how can you wrap it?

But how and why?

Because EnableCollectionSynchronization enables the CollectionView that sits in between the source collection and the ItemsControl to participate in synchronized access. So you need to call it on the actual INotifyCollectionChanged that you bind to in the view.

CollectionViewSource.GetDefaultView(Collection) is not equal to CollectionViewSource.GetDefaultView(this). In fact, Collection should be a private field of ThreadedObservableCollection<T> because it's an implementation detail. The view and WPF binds to an instance of ThreadedObservableCollection<T> and knows nothing about the internal ObservableCollection<T>.

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