Q:

Getting unity to resolve multiple instances of the same type

I want to do a simple resolve of multiple type registrations (ultimately constructor injected, but using .Resolve to see if Unity is even capable of such things.
In every case below, Unity resolves 0 items where it should be resolving 2.
Is there some switch in unity that turns on post-2007 behavior?  Or am I just drastically missing something?
Here is my code:
public interface IFoo {}
public class Foo1 : IFoo{}
public class Foo2 : IFoo{}

class Program
{
    static void Main(string[] args)
    {
        var container = new UnityContainer();
        container.RegisterType<IFoo, Foo1>();
        container.RegisterType<IFoo, Foo2>();

        // container.Resolve<IEnumerable<IFoo>>();   returns 0
        // container.ResolveAll<IFoo>(); returns 0

        var foos = container.Resolve<IFoo[]>();
        Console.WriteLine(foos.Count());

        Console.ReadLine();

    }
}

A:

In Unity there can only be one default registration (A registration without a name as in container.RegisterType<IFoo, Foo1>();
). If multiple default registrations are performed, the last one wins.
In order to register multiple implementation for the same interface, you need to assign names to those registrations:
container.RegisterType<IFoo, Foo1>("registration1");
container.RegisterType<IFoo, Foo2>("registration2");

Also, Unity only understand arrays by default. If you want to resolve as an array then you will be fine with the default behaviour. Otherwise you will need to register a mapping between the array and the collection you are interested in, like:
container.RegisterType<IEnumerable<IFoo>, IFoo[]>();

Another important note is that the default registration won't be returned when resolving a collection.
For example given:
container.RegisterType<IFoo, Foo1>();
container.RegisterType<IFoo, Foo2>("registration1");
container.RegisterType<IFoo, Foo3>("registration2");
container.RegisterType<IEnumerable<IFoo>, IFoo[]>();

If you resolve IEnumerable<IFoo>, the result will only contain instances of Foo2 and Foo3, but there will not be an instance of Foo1 because the default registration is not included.

