如何解决“未处理的异常类型ClassNotFoundException”和“向量是原始类型”错误？

发布于 2024-11-08 04:39:16 字数 538 浏览 1 评论 0原文

在 servlet 方面，我试图接收从小程序发送的向量。代码就像

InputStream in = request.getInputStream();
ObjectInputStream inputFromApplet = new ObjectInputStream(in);
Vector v = (Vector) inputFromApplet.readObject();

但是编译器指定以下代码是错误的。

Vector v = (Vector) inputFromApplet.readObject();

错误信息是

该行有多个标记
未处理的异常类型 ClassNotFoundException
Vector 是原始类型。对泛型类型 Vector 的引用应该参数化
Vector 是原始类型。对泛型类型 Vector 的引用应该参数化

我的代码有什么问题吗？谢谢。

原文

In servlet side, I am trying to receive an vector sent from the applet. The code is like

InputStream in = request.getInputStream();
ObjectInputStream inputFromApplet = new ObjectInputStream(in);
Vector v = (Vector) inputFromApplet.readObject();

But the compiler specifies that the following one is wrong.

Vector v = (Vector) inputFromApplet.readObject();

The error message is

Multiple markers at this line
Unhandled exception type ClassNotFoundException
Vector is a raw type. References to generic type Vector
should be parameterized
Vector is a raw type. References to generic type Vector
should be parameterized

What's wrong with my code? Thanks.

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煞人兵器 2024-11-15 04:39:17

对于例外情况，请将您的代码包装在

try {
   <your code here>
} catch (ClassNotFoundException e) {
   <code that happens when class can't be found>
}

其他消息中是警告，不会阻止您的程序编译。他们指的是需要参数化类型上的参数来遵守新规范（从 Java 1.5 开始）。因此，如果您想在 Vector 中存储整数，请使用 Vector。这会将方法更改为需要 Integers 而不是您通常获得的通用 Object，并提高类型安全性。

For the exception, wrap your code in

try {
   <your code here>
} catch (ClassNotFoundException e) {
   <code that happens when class can't be found>
}

The other messages are warnings which won't stop your program from compiling. They refer to needing a parameter on parametrised types to adhere to new specifications (as of Java 1.5). So, if you want to store integers in a Vector, use Vector<Integer>. This changes the methods to requiring Integers instead of the generic Object you'd normally get and increases type-safeness.

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