CODEWITHSUNDEEP
castle-windsor-3
Diana
Diana

Reputation: 839

How do I register default interfaces with Castle Windsor given an interface ancestor?

I have the following:

interface IAncestor { }

interface IDescendant1 : IAncestor { }

interface IDescendant2 : IAncestor { } 

class Descendant1 : IDescendant1 { }

class Descendant2 : IDescendant2 { }

What I would like to be able to do is automatically have Castle Windsor find all IDescendantX-DescendantX pairs without me having to specify them manually. Is this possible?

I've tried:

        container.Register(
            Classes.FromThisAssembly()
            .BasedOn<IAncestor>()
            .WithService.DefaultInterfaces()
            .LifestyleTransient()
        );

but this does not find the default interfaces. (I'm having trouble phrasing my question with the right terminology, so could not find a topic on SO that already answers this, sorry if it's a duplicate...)

Upvotes: 0

Views: 221

Answers (1)

jwalls
jwalls

Reputation: 16

Think the problem here is lack of access modifiers. If you add IncludeNonPublicTypes() the following test passes:

[Test]
public void Test() 
{        
    //Arrange
    WindsorContainer sut = new WindsorContainer();

    //Act
    sut.Register(
            Classes.FromThisAssembly()
            .IncludeNonPublicTypes()
            .BasedOn<IAncestor>()
            .WithService.DefaultInterfaces()
            .LifestyleTransient());

    //Assert
    Assert.That(sut.Resolve<IDescendant1>(), Is.InstanceOf<Descendant1>());
    Assert.That(sut.Resolve<IDescendant2>(), Is.InstanceOf<Descendant2>());
}

Upvotes: 0

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