Castle windsor: How to register two services with one instance of implementation?

Question

How to register two services with one instance of implementation? I used:

 _container.Register(Component.For(new [] { typeof(IHomeViewModel), typeof(IPageViewModel) }).
            ImplementedBy(typeof(HomeViewModel)).Named("IHomeViewModel").LifeStyle.Singleton)

But upper code registers two instances of HomeViewModel.

Answer 1

That's exactly the way to do it. See "Type Forwarding" in the docs. It registers one logical component accessible via IHomeViewModel or IPageViewModel. The following test passes:

public interface IHomeViewModel {}
public interface IPageViewModel {}
public class HomeViewModel: IHomeViewModel, IPageViewModel {}

[Test]
public void Forward() {
    var container = new WindsorContainer();
    container.Register(Component.For(new[] {typeof (IHomeViewModel), typeof (IPageViewModel)})
        .ImplementedBy(typeof(HomeViewModel)).Named("IHomeViewModel").LifeStyle.Singleton);
    Assert.AreSame(container.Resolve<IHomeViewModel>(), container.Resolve<IPageViewModel>());
}

BTW you might want to use generics instead of all those typeof, and also remove the lifestyle declaration, since singleton is the default:

container.Register(Component.For<IHomeViewModel, IPageViewModel>()
                            .ImplementedBy<HomeViewModel>());