Java Integer.parseInt()不适用于大数字 [英] Java Integer.parseInt() Not Working for Large Numbers

查看:951
本文介绍了Java Integer.parseInt()不适用于大数字的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我有以下简单的代码,用于检测给定的IPv4地址确实只有数值(即在剥离点后):

I have the following simple piece of code which is intended to detect that a given IPv4 address indeed only has numeric values (that is after the dots have been stripped):

import edu.gcc.processing.exceptions.net.IPAddressNumericException;

//Get the IP address
  String address = "239.255.255.255";

//Check to see if this is a number      
  try {
    String IPNumbers = address.replace(".", "");
    Integer.parseInt(IPNumbers);            
  } catch (NumberFormatException e) {
    System.out.print(e.getMessage());
  }

出于某种原因, NumberFormatException 被解雇,我收到此错误:

For some reason, the NumberFormatException is fired, and I get this error:

For input string: "239255255255"

有人可以帮我理解这个吗? parseInt()方法适用于较小的数字,例如 127001

Could someone please help me understand this? The parseInt() method works on smaller numbers, such as 127001.

感谢您的时间。

推荐答案

尝试使用 Long.parseLong(IPNumbers)

这篇关于Java Integer.parseInt()不适用于大数字的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

查看全文
登录 关闭
扫码关注1秒登录
发送“验证码”获取 | 15天全站免登陆