Using the Simple Injector, is it possible to get a singleton by its implementation type?

Question

If I register in the container something like:

container.Register<IShell, ShellViewModel>(Lifestyle.Singleton);

Is there a way to get the same instance using the "implementation type" ShellViewModel?

Example:

container.GetInstance<ShellViewModel>();

The above line returns an instance different from container.GetInstance<IShell>(). How can I make sure the instance is the same for both calls?

I solve it using ResolveUnregisteredType event.

private void ContainerResolveUnregisteredType(
    object sender, UnregisteredTypeEventArgs e)
{
    var producer = container.GetRootRegistrations()
        .FirstOrDefault(r => r.Registration
            .ImplementationType == e.UnregisteredServiceType);
    if (producer != null && producer.Lifestyle == Lifestyle.Singleton)
    {
        var registration = producer.Lifestyle
            .CreateRegistration(
                e.UnregisteredServiceType,
                producer.GetInstance,
                container);
        e.Register(registration);
    }
}

Is it the correct way?

Answer 1

You simply register them both as singleton:

container.RegisterSingleton<ShellViewModel>();
container.RegisterSingleton<IShell, ShellViewModel>();

UDPATE

Confirmed working with a simple unit test:

[TestMethod]
public void RegisterSingleton_TwoRegistrationsForTheSameImplementation_ReturnsTheSameInstance()
{
    var container = new Container();

    container.RegisterSingleton<ShellViewModel>();
    container.RegisterSingleton<IShell, ShellViewModel>();

    var shell1 = container.GetInstance<IShell>();
    var shell2 = container.GetInstance<Shell>();

    Assert.AreSame(shell1, shell2);
}