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

问题描述

鉴于此代码段可以轻松粘贴到Linqpad中（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 的输出显示狗，虽然作为IAnimal返回，但is是一个ISmartAnimal。 （对于两个对象都相同）

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？ True

Type: Animal
Is smart? True

这是因为 GetType（）返回对象的具体类型，而不是它当前的界面。

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是否可以使它 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：

这似乎是一项学术活动。让一个类（Animal）实现一个你不想总是暴露的接口（ISmartAnimal）似乎很奇怪。但我问，因为我在Entity Framework中遇到过类似的东西。如果你想要，你可以在这里阅读它，但它会转向EF-具体功能。如果您不想深入研究，那么 Animal 必须实现两个接口。

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：

与真实动物有任何相似之处纯属巧合:)

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

推荐答案

听起来你对编译时类型的 dog 变量感兴趣。你可以通过 ReportTypeProperties generic来让排序得到这个，并让编译器根据变量的类型推断出类型：

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，这可以通过各种方式进行游戏，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 - 感觉不太可能我朝这个方向前进将会带来一个好的设计。

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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