如何判断方法返回的是哪个接口 [英] How to tell which interface is returned by a method

问题描述

鉴于此代码片段可以轻松粘贴到 Linqpad(或在 Visual Studio 控制台解决方案中稍作修改):

Given this code snippet that can be readily pasted into Linqpad (or slightly modified in a Visual Studio console solution):

void Main()
{
    var cat = this.GetCat();
    var dog = this.GetDog();
    cat.Think();
    cat.ThinkHarder();
    //dog.Think(); // Does not compile.
    //dog.ThinkHarder(); // Does not compile.

    //if ([dog is returned as ISmartAnimal]) // What to put here?
        ((ISmartAnimal)dog).Think(); // Compiles, runs, but shouldn't.

    reportTypeProperties(cat);
    reportTypeProperties(dog);
}

interface IAnimal
{
    string Name { get; set; }
}

interface ISmartAnimal : IAnimal
{
    void Think();
}

class Animal : IAnimal, ISmartAnimal
{
    public string Name { get; set; }
    public void Think() { }
}

ISmartAnimal GetCat()
{
    return new Animal();
}

IAnimal GetDog()
{
    return new Animal();
}

static void reportTypeProperties(object obj)
{
    var type = obj.GetType();
    Console.WriteLine("Type: {0}", type.Name);
    Console.WriteLine("Is smart? {0}", obj is ISmartAnimal);
}

static class ext
{
    public static void ThinkHarder(this ISmartAnimal animal)
    { }
}

reportTypeProperties 的输出表明 dog 虽然返回为 IAnimal，但是"一个 ISmartAnimal.(两个对象相同)

The output of reportTypeProperties shows that dog, although returned as IAnimal, "is" an ISmartAnimal. (Same for both objects)

类型:动物
聪明吗?真

Type: Animal
Is smart? True

这是因为 GetType() 返回对象的具体类型，而不是其当前接口.

This is because GetType() returns the concrete type of the object, not its current interface.

我的问题.有没有办法告诉 dog 作为 IAnimal 返回?(参见伪代码).编译器知道(quickview 也是如此).假设我有一些动物对象，我想在运行时代码中检查我是否可以Think().

My question. Is there a way to tell that dog is returned as IAnimal? (see pseudocode). The compiler knows (so does quickview). Suppose I had some animal object and I wanted to inspect in runtime code whether or not I can make it Think().

背景:
这似乎是一种学术练习.让一个类 (Animal) 实现一个您不想总是公开的接口 (ISmartAnimal) 似乎很奇怪.但是我问是因为我在实体框架中遇到了类似的事情.如果你愿意，你可以在这里阅读它，但它转向EF-具体功能.如果您不想深入研究，只需说 Animal 必须实现这两个接口就足够了.

Background:
This may seem an academic exercise. And it may seem strange to have a class (Animal) implement an interface (ISmartAnimal) that you don't want to expose always. But I ask because I encountered something similar in Entity Framework. If you want you can read about it here, but it diverts to EF-specific features. If you don't want to delve into that it suffices to say that it's imperative that Animal implement both interfaces.

免责声明:
与真实动物的任何相似之处纯属巧合:)"

Disclaimer:
"Any resemblance to real animals is purely coincidental :)"

推荐答案

听起来您对 dog 变量的编译时类型很感兴趣.您可以排序得到这个，通过使 ReportTypeProperties 泛型并让编译器根据变量的类型推断类型:

It sounds like you're interested in the compile-time type of the dog variable. You can sort of get this, by making ReportTypeProperties generic and letting the compiler infer the type based on the variable's type:

static void ReportTypeProperties<T>(T obj)
{
    Console.WriteLine("Compile-time type: {0}", typeof(T).Name);
    Console.WriteLine("Actual type: {0}", obj.GetType().Name);
    Console.WriteLine("Is smart? {0}", obj is ISmartAnimal);
}

请注意，这可以通过多种方式进行游戏，例如

Note that this can be gamed in various ways, e.g.

object dog = GetDog();
ReportTypeProperties(dog); // Would show as object

或

IAnimal dog = GetDog();
ReportTypeProperties<object>(dog); // Would show as object

目前还不清楚这里的大局是什么 - 我觉得朝这个方向发展不太可能会带来好的设计.

It's not really clear what the bigger picture is here - it feels unlikely to me that going in this direction is going to lead to a good design.

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