温莎城堡如何响应实现多个接口的类？

问题描述

例如，我有两个接口：ICustomerService和IOrderService，每个接口都有几个函数，例如GetCustomer，GetOrder等。

For example I have two interfaces: ICustomerService and IOrderService which each has a couple of functions like GetCustomer, GetOrder, etc.

我希望一个类实现两个接口：服务器。

I want one class to implement both interfaces: Server.

温莎城堡对此有何反应？
首先有可能吗？
当我基于两个接口之一解析Server对象时，会得到相同的对象吗？
当我有一个在其参数中同时包含两个接口的构造函数时，会发生什么？仍然会构造一个对象。

How does Castle Windsor respond to this?Is it possible in the first place?When I resolve the Server object based on one of the two interfaces, will I get the same object?What happens when I have a constructor that has both interfaces in its parameters? Will there still be one object constructed.

假定生活方式保留为默认值：Singleton。

assuming the LifeStyle is left to its default: Singleton.

推荐答案

在CLR类型和Windsor服务或组件之间（）。

There's no hard one-to-one mapping between a CLR type and a Windsor service or component (glossary if needed is here).

因此，您可以使用以下情况：

So you can have any of the following scenarios:

• 许多具有不同实现类型的组件公开单个服务

• Many Components with different implementation types expose a single Service

container.Register(
   Component.For<IFoo>().ImplementedBy<Foo1>(),
   Component.For<IFoo>().ImplementedBy<Foo2>()
);

许多具有相同实现类型的组件公开单个服务

Many Components with same implementation type expose a single Service

container.Register(
   Component.For<IFoo>().ImplementedBy<Foo1>(),
   Component.For<IFoo>().ImplementedBy<Foo1>().Named("second")
);

许多具有相同实现类型的组件公开了不同的服务

Many Components with same implementation type expose different Services

container.Register(
   Component.For<IFoo>().ImplementedBy<Foo1>(),
   Component.For<IBar>().ImplementedBy<Foo1>().Named("second")
);

单个组件公开不同的服务

Single Component expose different Services

container.Register(
   Component.For<IFoo, Bar>().ImplementedBy<Foo1>()
);

您可以看到，是的-是可能，以及是否会获得相同的实例（假设单例）取决于您选择的选项-是否两个服务都将由同一组件公开。

As you can see, yes - it is possible, and whether or not you'll get the same instance (assuming singleton) depends on which option you chose - if both Services will be exposed by the same Component, or not.

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