抛出一个UnsupportedOperationException异常 [英] Throwing an UnsupportedOperationException

问题描述

所以一个方法描述如下：

So one of the method descriptions goes as follows:

public BasicLinkedList addToFront（T data）此操作无效
为排序列表。将使用消息排序列表的无效操作
生成UnsupportedOperationException。

public BasicLinkedList addToFront(T data) This operation is invalid for a sorted list. An UnsupportedOperationException will be generated using the message "Invalid operation for sorted list."

我的代码如下所示：

public BasicLinkedList<T> addToFront(T data) {
    try {
        throw new UnsupportedOperationException("Invalid operation for sorted list.");
    } catch (java.lang.UnsupportedOperationException e) {
        System.out.println("Invalid operation for sorted list.");
    }
    return this;
}

这是正确的做法吗？我刚刚使用 println（）打印出消息，但有不同的方式来生成消息？

Is this the right way of doing this? I just printed out the message using println() but is there a different way to generate the message?

推荐答案

您不想在方法中捕获异常 - 重点是让调用者知道不支持该操作：

You don't want to catch the exception in your method - the point is to let callers know that the operation is not supported:

public BasicLinkedList<T> addToFront(T data) {
    throw new UnsupportedOperationException("Invalid operation for sorted list.");
}

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