英文:
Why casting to unrelated interface produce ClassCastException instead of compile time exception?
问题
为什么在这个代码片段中，尽管Tree接口与Bug类无关，但它不会产生编译时异常？
interface Tree {}
class Bug {
public static void main(String[] args) {
Bug bug = new Bug();
Tree tree = (Tree) bug;
}
}
英文:
Why in this snippet while Tree interface is not related to Bug class it does not produce compile time exception?
interface Tree {}
class Bug {
public static void main(String[] args) {
Bug bug = new Bug();
Tree tree = (Tree) bug;
}
}
答案1
得分: 2
将 Bug 标记为 final，你将会得到预期的编译错误。
这是因为 Bug 可能有一个子类实际上实现了 Tree 接口。如果 Bug 实际上存储了对该子类实例的引用，那么类型转换将会成功。由于存在类型转换可能成功的情况，编译器不会阻止你进行类型转换。
在大多数情况下，你可以从任何非 final 类型转换到任何接口。根据 JLS §5.5.1，当你尝试将引用类型为 S 的变量转换为接口 T 时：
如果 S 不是 final 类型（§8.1.1），那么如果存在接口
T的超类型X，和变量S的超类型Y，使得X和Y都是可以证明不同的参数化类型，并且X和Y的擦除操作是相同的，那么在编译时会产生错误。否则，在编译时转换总是合法的（因为即使
S没有实现T，S的子类可能会实现）。
英文:
Mark Bug as final and you will get your compiler error as expected.
This is because Bug could have a subclass that does implement Tree. If bug actually stores a reference to an instance of that subclass, then the cast will succeed. Since there is a chance that the cast can succeed, the compiler doesn't stop you from casting.
In most cases, you can cast from any non-final class to any interface. According to the JLS §5.5.1, when you try to cast a variable of reference type S to interface T:
> 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.
>
> Otherwise, the cast is always legal at compile time (because even if S does not implement T, a subclass of S might).
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