try-catch为零除 [英] try-catch for division by zero

问题描述

我的问题是关于try-catch块的简单除法示例。你看到第一行尝试？如果我将这两个变量中的任何一个转换为double，则程序无法识别catch块。在我看来，我是否必须执行捕获块。这段代码有什么问题？

My question is about try-catch blocks on a simple division by zero example. You see the first line of try? If I cast any of those two variables to the double the program does not recognize the catch block. In my opinion, whether I cast or not only the catch block must be executed. What is wrong on this code?

public static void main(String[] args) {

    int pay=8,payda=0;  

    try {

        double result=pay/(double)payda; // if I cast any of the two variables, program does not recognize the catch block, why is it so?
        System.out.println(result);
        System.out.println("inside-try");

    } catch (Exception e) {

        System.out.println("division by zero exception");
        System.out.println("inside-catch");

    }
}

推荐答案

除以零对浮点数有效。

• 1/0产生无穷大。


• （ - 1）/ 0产生-Infinity。


• 0/0产生NaN。

这些数字在IEEE 754中正确定义。

These "numbers" are properly defined in IEEE 754.

另一方面，整数除以零会抛出，因为无法将无穷大表示为 int 。

Integer division by zero, on the other hand, throws because one cannot represent infinity as an int.

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