验证IP地址（带掩码） [英] Validate an IP Address (with Mask)

问题描述

我有IP地址和掩码，例如 10.1.1.1/32 。我想检查 10.1.1.1 是否在该范围内。是否有一个库或实用程序可以做到这一点，或者我需要自己写一些东西？ 解决方案

首先，您需要将您的IP地址转换为平面 int s，这将更易于使用：

  String s =10.1.1.99; 
 Inet4Address a =（Inet4Address）InetAddress.getByName（s）; 
 byte [] b = a.getAddress（）; 
 int i =（（b [0]& 0xFF）<24} | 
（（b [1]& 0xFF）<< 16）| 
（（b [2]& 0xFF）<< 8）| 
（（b [3]& 0xFF）<< 0）; 
  

将IP地址设为普通 int 你可以做一些算术来执行检查：

pre $ int $ subnet = 0x0A010100; // 10.1.1.0/24
int bits = 24;
int ip = 0x0A010199; // 10.1.1.99

//创建位掩码来清除不相关的位。对于10.1.1.0/24，这是
// 0xFFFFFF00 - 前24位是1，最后8位是0。
//
// -1 == 0xFFFFFFFF
// 32 - 位== 8
// -1 << 8 == 0xFFFFFF00
mask = -1<< （32位）

if（（subnet& mask）==（ip& mask））{
// IP地址在子网中。
}

I have ip addresses and a mask such as 10.1.1.1/32. I would like to check if 10.1.1.1 is inside that range. Is there a library or utility that would do this or do I need to write something myself?

解决方案

First you'll want to convert your IP addresses into flat ints, which will be easier to work with:

String       s = "10.1.1.99";
Inet4Address a = (Inet4Address) InetAddress.getByName(s);
byte[]       b = a.getAddress();
int          i = ((b[0] & 0xFF) << 24) |
                 ((b[1] & 0xFF) << 16) |
                 ((b[2] & 0xFF) << 8)  |
                 ((b[3] & 0xFF) << 0);

Once you have your IP addresses as plain ints you can do some bit arithmetic to perform the check:

int subnet = 0x0A010100;   // 10.1.1.0/24
int bits   = 24;
int ip     = 0x0A010199;   // 10.1.1.99

// Create bitmask to clear out irrelevant bits. For 10.1.1.0/24 this is
// 0xFFFFFF00 -- the first 24 bits are 1's, the last 8 are 0's.
//
//     -1        == 0xFFFFFFFF
//     32 - bits == 8
//     -1 << 8   == 0xFFFFFF00
mask = -1 << (32 - bits)

if ((subnet & mask) == (ip & mask)) {
    // IP address is in the subnet.
}

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