How to define the default implementation of an interface in c#?

Question

There is some black magic code in c# where you can define the default implementation of an interface.

So you can write

var instance = new ISomeInterf         


        

Answer 1

Only if ISomeInterface is a class.

Update (for clarification):

Jon Skeet has a talk where he mentions default implementations for interfaces. They are not part of the C# language, though. The talk is about what Jon Skeet would like to see in a future version of C#.

For now, the only default implementations are done via (possibly abstract) base classes.

Answer 2

Maybe you refer to Dependency Injection? Where when using DI framework (such as Ninject or Unity), you can define default instance for each interface and then using it like this:

(assuming you have IWeapon interface and Sword implements it)

IKernel kernel = new StandardKernel();
kernel.Bind<IWeapon>().To<Sword>();
var weapon = kernel.Get<IWeapon>();

But Ninject (and most other IoC frameworks) can do some more clever things, like: let's say we have the class Warrior that takes IWeapon as a parameter in its constructor. We can get an instance of Warrior from Ninject:

var warrior = kernel.Get<Warrior>();

Ninject will pass the IWeapon implementation we specified to the Warrior constructor method and return the new instance.

Other than that, I don't know of any in-language feature that allows this kind of behavior.

Answer 3

Here comes the black magic:

class Program
{
    static void Main()
    {
        IFoo foo = new IFoo("black magic");
        foo.Bar();
    }
}

[ComImport]
[Guid("C8AEBD72-8CAF-43B0-8507-FAB55C937E8A")]
[CoClass(typeof(FooImpl))]
public interface IFoo
{
    void Bar();
}

public class FooImpl : IFoo
{
    private readonly string _text;
    public FooImpl(string text)
    {
        _text = text;
    }

    public void Bar()
    {
        Console.WriteLine(_text);
    }
}

Notice that not only you can instantiate an interface but also pass arguments to its constructor :-)