C# - 动态关键字和接口实现

发布于 2024-08-31 16:50:50 字数 472 浏览 9 评论 0原文

我假设这是不可能的，但在进一步挖掘之前，有没有办法做这样的事情：

public void ProcessInterface(ISomeInterface obj) {}

//...

dynamic myDyn = GetDynamic<ISomeInterface>() 
ProcessInterface(myDyn);

我看到了 post 争论它，但听起来它没有包含在内。

一些上下文：通过 COM 公开的 .Net 程序集 -> Silverlight 应用程序使用接口实现类。通过接口引用对象会很好。真没想到，原来的目的是这样的……

原文

I'm assuming this isn't possible but before digging further is there a way to do something like this:

public void ProcessInterface(ISomeInterface obj) {}

//...

dynamic myDyn = GetDynamic<ISomeInterface>() 
ProcessInterface(myDyn);

I've seen a post arguing for it but it sounds like it wasn't included.

A little context: .Net assembly exposed through COM -> Silverlight app consuming interface-implementing classes. Would be nice to refer to the objects by interface. I really don't expect that this was what was intended...

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哽咽笑 2024-09-07 16:50:50

不，dynamic 不会使类型假装实现接口（即使它通过 dynamic 拥有所有方法）。将其传递给ProcessInterface本质上会消除动态。

动态 取决于调用代码，就像实现对象一样。更多，甚至。

然而，您可以创建一个使用鸭子类型的接口包装器：

class Foo : IBar {
    readonly dynamic duck;
    public Foo(dynamic duck) { this.duck = duck; }

    void IBar.SomeMethod(int arg) {
        duck.SomeMethod(arg);
    }
    string IBar.SomeOtherMethod() {
        return duck.SomeOtherMethod();
    }
}
interface IBar {
    void SomeMethod(int arg);
    string SomeOtherMethod();
}

No, dynamic won't make a type pretend to implement an interface (even if it has, via dynamic, all the methods). Passing it to ProcessInterface essentially takes away the dynamic.

dynamic depends on the calling code just as much as the implementing object. More, even.

You could however make an interface wrapper that uses duck typing:

class Foo : IBar {
    readonly dynamic duck;
    public Foo(dynamic duck) { this.duck = duck; }

    void IBar.SomeMethod(int arg) {
        duck.SomeMethod(arg);
    }
    string IBar.SomeOtherMethod() {
        return duck.SomeOtherMethod();
    }
}
interface IBar {
    void SomeMethod(int arg);
    string SomeOtherMethod();
}

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