捕获 ArrayOutOfBoundsException

Question

我正在为我的学校做一个项目，我已经处理了所有异常，但我就是无法处理 ArrayOutOfBoundsException。所以我在 Whosebug 上搜索了一个解决方案，它说使用 IndexOutOfBoundsException（link 到解决方案：why is it not catching the arrays out of bound?）但它没有用。请帮忙？这是我的代码：

for (int i = 0; i < 4; i++) {
    System.out.println(fruit[i] + fruitprice[i]);
}
System.out.println("\n" + "please enter the number marked for the icecream you want to buy");
do {
    x = 0;
    try {
        icenum1 = Integer.parseInt( in .readLine()) - 1;
    } catch (NumberFormatException n) {
        System.out.println("please enter in numbers and not in words.... please try again");
        x = 1;
    } catch (IndexOutOfBoundsException e) {
        System.out.println("please enter a number out of the given numbers and not any other number");
        x = 1;
    }
} while (x != 0);
System.out.print("\u000c");
System.out.println("your choice is " + fruit[icenum1]);

Answer 1

如果它仍然符合您的代码逻辑，请将这些行移动到您的 try 块中。

System.out.print("\u000c");
System.out.println("your choice is " + fruit[icenum1]);

如果抛出 IndexOutOfBoundException，它将从第二行开始，因为它会尝试访问当前数组中不存在的索引。

捕获 ArrayOutOfBoundsException

catching ArrayOutOfBoundsException

java

exception-handling