处理 ArrayStoreException [英] Dealing with an ArrayStoreException

问题描述

Object[] o = "a;b;c".split(";");
o[0] = 42;

投掷

java.lang.ArrayStoreException: java.lang.Integer

同时

String[] s = "a;b;c".split(";");
Object[] o = new Object[s.length];
for (int i = 0; i < s.length; i++) {
    o[i] = s[i];
}
o[0] = 42;

没有.

有没有其他方法可以在不创建临时 String[] 数组的情况下处理该异常?

Is there any other way to deal with that exception without creating a temporary String[] array?

推荐答案

在 Java 中，数组也是一个对象.

In Java an array is also an object.

您可以将子类型的对象放入超类型的变量中.例如，您可以将 String 对象放入 Object 变量中.

You can put an object of a subtype into a variable of a supertype. For example you can put a String object into an Object variable.

不幸的是，Java 中的数组定义以某种方式被破坏了.String[] 被认为是 Object[] 的子类型，但这是错误的！有关更详细的解释，请阅读协变和逆变"，但其本质是:只有当子类型满足超类型的所有义务时，该类型才应被视为另一种类型的子类型.这意味着，如果您获得的是子类型对象而不是超类型对象，则不应期望与超类型约定相矛盾的行为.

Unfortunately, the array definition in Java is somehow broken. String[] is considered a subtype of Object[], but that is wrong! For a more detailed explanation read about "covariance and contravariance", but the essence it this: A type should be considered a subtype of another type only if the subtype fulfills all obligations of the supertype. That means, that if you get a subtype object instead of a supertype object, you should not expect behavior contradictory to supertype contract.

问题是String[] 只支持Object[] 合约的part.例如，您可以从 Object[] 中读取 Object 值.您还可以从 String[] 中读取 Object 值(恰好是 String 对象).到现在为止还挺好.问题出在合同的另一部分.你可以把any Object 放入Object[].但是你不能把any Object 放入String[].因此，String[] 不应被视为 Object[] 的子类型，但 Java 规范说它是.因此，我们会产生这样的后果.

Problem is that String[] only supports a part of Object[] contract. For example you can read Object values from Object[]. And you can also read Object values (which happen to be String objects) from String[]. So far so good. Problem is with the other part of contract. You can put any Object into Object[]. But you cannot put any Object into String[]. Therefore, String[] should not be considered a subtype of Object[], but Java specification says it is. And thus we have consequences like this.

(注意泛型类又出现了类似的情况，不过这次正确地解决了.List is not> List