将对象强制转换为未实现的接口 [英] Casting an object to an interface which is not implemented

问题描述

我在学习书中发现以下问题，有些困惑:

I found following question in my study book and a bit confused:

给出以下代码，该选项(如果用于替换/* INSERT CODE HERE */)将使类型为Roamable的引用变量引用该对象. Phone类? (选择1个选项.)

Given the following code, which option, if used to replace /* INSERT CODE HERE */, will enable a reference variable of type Roamable to refer to an object of the Phone class? (Select 1 option.)

interface Roamable{}
class Phone {}
class Tablet extends Phone implements Roamable {
    //INSERT CODE HERE
}

选项是:

1. Roamable var = new Phone();
2. Roamable var = (Roamable)Phone();
3. Roamable var = (Roamable)new Phone();
4. 因为接口Roamable和类Phone不相关，所以引用变量 类型为Roamable的对象不能引用类为Phone的对象.

1. Roamable var = new Phone();
2. Roamable var = (Roamable)Phone();
3. Roamable var = (Roamable)new Phone();
4. Because interface Roamable and class Phone are unrelated, a reference variable of type Roamable can’t refer to an object of class Phone.

我认为正确的选项是4，但是它说是3.

I thought the correct option is 4, however it says it is 3.

但是，Phone没有实现Roamable接口，所以您不能进行投射，可以吗?

But, Phone doesn't implement Roamable interface, so you can't cast, can you?

推荐答案

正确的答案是3:编译器仅看到将Phone强制转换为Roamable，并且Phone不是最终的，因此它认为正在转换的对象(虽然称为Phone)可能是做实现Roamable的Phone的子类，因此不会发出编译时错误或警告.

The correct answer is 3: The compiler sees only that a Phone is being cast to a Roamable and that Phone is not final, so it thinks that the object being cast, although referred to as Phone may be a subclass of Phone that does implement Roamable, so no compile-time error or warning is issued.

根据 JLS第5章

5.5.1.引用类型转换

给定编译时参考类型S(源)和编译时参考类型T(目标)，如果由于以下规则而没有发生编译时错误，则存在从S到T的转换转换. 如果T是接口类型:

Given a compile-time reference type S (source) and a compile-time reference type T (target), a casting conversion exists from S to T if no compile-time errors occur due to the following rules. If T is an interface type:

如果S不是最终类(第8.1.1节)，则如果存在T的超类型X和S的超类型Y，则X和Y都是可证明是截然不同的参数化类型，并且X和Y的擦除相同，则发生编译时错误.

If S is not a final class (§8.1.1), then, if there exists a supertype X of T, and a supertype Y of S, such that both X and Y are provably distinct parameterized types, and that the erasures of X and Y are the same, a compile-time error occurs.

否则，强制转换在编译时始终是合法的(因为即使S不实现T，S的子类也可能是这样).

Otherwise, the cast is always legal at compile time (because even if S does not implement T, a subclass of S might).

如果S是最终类(第8.1.1节)，则S必须实现T，否则会发生编译时错误.

If S is a final class (§8.1.1), then S must implement T, or a compile-time error occurs.

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以下代码编译:

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The following code compiles:

interface Roamable{}
class Phone {}
class Tablet extends Phone implements Roamable {
    Roamable var = (Roamable)new Phone(); // Compiles
}

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