如何在 Java 中将 IP 地址范围转换为 CIDR？

Question

我正在尝试将 IP 地址范围转换为 Java 中的 CIDR 表示法。有人可以举例说明如何实现这一点吗？

我使用 SubnetUtils 将 CIDR 转换为 IP 地址范围，但反过来我无法弄清楚。

例如：（使用http://ip2cidr.com/）

输入 1： 5.10.64.0
输入 2： 5.10.127.255
结果： 5.10.64.0/18

谢谢， 达瓦尔

Answer 1

import java.util.ArrayList;
import java.util.List;

public class RangeToCidr {
    public static List<String> range2cidrlist( String startIp, String endIp ) {
        // check parameters
        if (startIp == null || startIp.length() < 8 ||
            endIp == null || endIp.length() < 8) return null;
        long start = ipToLong(startIp);
        long end = ipToLong(endIp);
        // check parameters
        if (start > end) return null;

        List<String> result = new ArrayList<String>();
        while (start <= end) {
            // identify the location of first 1's from lower bit to higher bit of start IP
            // e.g. 00000001.00000001.00000001.01101100, return 4 (100)
            long locOfFirstOne = start & (-start);
            int maxMask = 32 - (int) (Math.log(locOfFirstOne) / Math.log(2));

            // calculate how many IP addresses between the start and end
            // e.g. between 1.1.1.111 and 1.1.1.120, there are 10 IP address
            // 3 bits to represent 8 IPs, from 1.1.1.112 to 1.1.1.119 (119 - 112 + 1 = 8)
            double curRange = Math.log(end - start + 1) / Math.log(2);
            int maxDiff = 32 - (int) Math.floor(curRange);

            // why max?
            // if the maxDiff is larger than maxMask
            // which means the numbers of IPs from start to end is smaller than mask range
            // so we can't use as many as bits we want to mask the start IP to avoid exceed the end IP
            // Otherwise, if maxDiff is smaller than maxMask, which means number of IPs is larger than mask range
            // in this case we can use maxMask to mask as many as IPs from start we want.
            maxMask = Math.max(maxDiff, maxMask);

            // Add to results
            String ip = longToIP(start);
            result.add(ip + "/" + maxMask);
            // We have already included 2^(32 - maxMask) numbers of IP into result
            // So the next round start must add that number
            start += Math.pow(2, (32 - maxMask));
        }
        return result;
    }

    private static long ipToLong(String strIP) {
        String[] ipSegs = strIP.split("\\.");
        long res = 0;
        for (int i = 0; i < 4; i++) {
            res += Long.valueOf(ipSegs[i]) << (8 * (3 - i));
        }
        return res;
    }

    private static String longToIP(long longIP) {
        StringBuffer sb = new StringBuffer();
        sb.append(longIP >>> 24).append(".")
          .append((longIP & 0x00FFFFFF) >>> 16).append(".")
          .append(String.valueOf((longIP & 0x0000FFFF) >>> 8)).append(".")
          .append(String.valueOf(longIP & 0x000000FF));

        return sb.toString();
    }
}

Answer 2

open-source IPAddress Java library 可以为您做到这一点。免责声明：我是 IPAddress 库的项目经理。

这是执行此操作的示例代码：

static void toPrefixBlocks(String str1, String str2) {
    IPAddressString string1 = new IPAddressString(str1);
    IPAddressString string2 = new IPAddressString(str2);
    IPAddress one = string1.getAddress(), two = string2.getAddress();
    IPAddressSeqRange range = one.toSequentialRange(two);
    System.out.println("starting with range " + range);
    IPAddress blocks[] = range.spanWithPrefixBlocks();
    System.out.println("prefix blocks are " + Arrays.asList(blocks));
}

这是您将如何使用 IPAddress 从原始 CIDR 字符串生成范围：

static String[] toRange(String str) {
    IPAddressString string = new IPAddressString(str);
    IPAddress addr = string.getAddress();
    System.out.println("starting with CIDR " + addr);
    IPAddress lower = addr.getLower(), upper = addr.getUpper();
    System.out.println("range is " + lower + " to " + upper);
    return new String[] {lower.toString(), upper.toString()};
}

对于您的示例，我们运行以下代码：

String strs[] = toRange("5.10.64.0/18");
System.out.println();
toPrefixBlocks(strs[0], strs[1]);

输出是：

starting with CIDR 5.10.64.0/18
range is 5.10.64.0/18 to 5.10.127.255/18

starting with range 5.10.64.0 -> 5.10.127.255
prefix blocks are [5.10.64.0/18]

Answer 3

如果你还没有从我的 cmets 中弄清楚：

IP 数学必须以二进制形式完成。 IP 地址和掩码是无符号整数（IPv4 为 32 位，IPv6 为 128 位）。您只需要知道地址和掩码，就可以弄清楚其他所有内容。

这是您想要完成的算法，它适用于 IPv4 和 IPv6。

根据您的问题，您将获得子网（输入 1）和最后一个地址（输入 2）。

1. 从无符号整数中减去输入 1 的无符号整数 输入 2。结果是反子网掩码。逆 子网掩码必须是 0，或者反向子网掩码加上 1 必须是 2 的幂，否则其中一个输入有错误（STOP， 输入错误）。
2. 反向掩码的NOT（步骤 1 的结果）是子网掩码。
3. 如果输入 1 AND 子网掩码不等于输入 1，则您有一个 输入之一错误（STOP，INPUT ERROR）。
4. 掩码长度（CIDR编号）是1位在 子网掩码。有几种方法可以计算1的数量 二进制数中的位，但如果子网掩码是最大值 整数（或反掩码为0），则掩码长度为32 (IPv4) 或128 (IPv6)。你可以循环，计算循环次数 并将子网掩码向左移动，直到等于0，循环 计算循环次数并将反向掩码移动到 直到等于0 然后将1 添加到总数中 从32 (IPv4) 或128 (IPv6) 中减去总数，或减去 总逆掩码的2 的幂的指数 加上 1 来自 32 (IPv4) 或 128 (IPv6)。
5. 此时，您已验证输入 1（子网）、输入 2（最后 地址），并计算掩码长度（CIDR 编号）。
6. 最终结果将是<Input 1>/<Mask Length>。

你的例子：

第 1 步 (5.10.127.255 - 5.10.64.0 = 0.0.64.127)：

101000010100111111111111111 - 01000010100100000000000000 = 11111111111111

第 2 步（NOT 0.0.64.255 = 255.255.192.0 是 2 的幂）：

NOT 00000000000000000011111111111111 = 11111111111111111100000000000000

第 3 步 (5.10.64.0 AND 255.255.192.0 = 5.10.64.0)：

01000010100100000000000000 AND 11111111111111111100000000000000 = 01000010100100000000000000

第 4 步 (0.0.64.255 + 1 = 0.0.65.0 = 2^14, exponent of 2^14 = 14, 32 - 14 = 18)：

00000000000000000011111111111111 + 1 = 00000000000000000100000000000000 = 2^14, exponent of 2^14 = 14, 32 - 14 = 18

第 5 步（输入 1 = 5.10.64.0，输入 2 = 5.10.127.255，掩码长度 = 18）

第 6 步（最终结果 = 5.10.64.0/18）

Answer 4

所以，我可以在这里找到 Java 代码：In Java, given an IP Address range, return the minimum list of CIDR blocks that covers the range

public class IP2CIDR {

    public static void main(String[] args) {
        System.out.println(range2cidrlist("5.104.109.160", "5.104.109.191"));
    }

    public static List<String> range2cidrlist( String startIp, String endIp ) {         
        long start = ipToLong(startIp);         
        long end = ipToLong(endIp);           

        ArrayList<String> pairs = new ArrayList<String>();         
        while ( end >= start ) {             
            byte maxsize = 32;             
            while ( maxsize > 0) {                 
                long mask = CIDR2MASK[ maxsize -1 ];                 
                long maskedBase = start & mask;                 

                if ( maskedBase != start ) {                     
                    break;                 
                }                 

                maxsize--;             
            }               
            double x = Math.log( end - start + 1) / Math.log( 2 );             
            byte maxdiff = (byte)( 32 - Math.floor( x ) );             
            if ( maxsize < maxdiff) {                 
                maxsize = maxdiff;             
            }             
            String ip = longToIP(start);             
            pairs.add( ip + "/" + maxsize);             
            start += Math.pow( 2, (32 - maxsize) );         
        }         
        return pairs;     
    }       

    public static final int[] CIDR2MASK = new int[] { 0x00000000, 0x80000000,             
        0xC0000000, 0xE0000000, 0xF0000000, 0xF8000000, 0xFC000000,             
        0xFE000000, 0xFF000000, 0xFF800000, 0xFFC00000, 0xFFE00000,             
        0xFFF00000, 0xFFF80000, 0xFFFC0000, 0xFFFE0000, 0xFFFF0000,             
        0xFFFF8000, 0xFFFFC000, 0xFFFFE000, 0xFFFFF000, 0xFFFFF800,             
        0xFFFFFC00, 0xFFFFFE00, 0xFFFFFF00, 0xFFFFFF80, 0xFFFFFFC0,             
        0xFFFFFFE0, 0xFFFFFFF0, 0xFFFFFFF8, 0xFFFFFFFC, 0xFFFFFFFE,             
        0xFFFFFFFF };       

    private static long ipToLong(String strIP) {         
        long[] ip = new long[4];         
        String[] ipSec = strIP.split("\\.");         
        for (int k = 0; k < 4; k++) {             
            ip[k] = Long.valueOf(ipSec[k]);         
        }         

        return (ip[0] << 24) + (ip[1] << 16) + (ip[2] << 8) + ip[3];     
    }       

    private static String longToIP(long longIP) {         
        StringBuffer sb = new StringBuffer("");         
        sb.append(String.valueOf(longIP >>> 24));         
        sb.append(".");         
        sb.append(String.valueOf((longIP & 0x00FFFFFF) >>> 16));         
        sb.append(".");         
        sb.append(String.valueOf((longIP & 0x0000FFFF) >>> 8));         
        sb.append(".");         
        sb.append(String.valueOf(longIP & 0x000000FF));   

        return sb.toString();     
    }
}

感谢大家的见解和帮助！

Answer 5

Python 中的短小精悍：

#!/usr/bin/env python
import ipaddress
import math

ip_from = '5.10.64.0'
ip_to = '5.10.127.255'
ip_from_long = int(ipaddress.ip_address(unicode(ip_from)))
ip_to_long = int(ipaddress.ip_address(unicode(ip_to)))
ip_range = ip_to_long - ip_from_long
ip_range +=1
# the clever line of code
cidr_range = math.log(4294967296/ip_range)/math.log(2)
# test for a zero/non-zero fractional part
if cidr_range % 1 == 0:
  # the output will be: 5.10.64.0/18
  print ip_from + '/' + str(int(cidr_range))
else:
  print "Error: Not an exact CIDR range - " + str(cidr_range)

Answer 6

public static int log2(int i) {
    int count = 0;
    i >>= 1;
    while(i > 0) {
        count++;
        i >>= 1;
    }
    return count;
}

public static List<String> range2CIDR(String startIp, String endIp) {
    List<String> res = new ArrayList<>();
    try {
        int start = ipS2I(startIp);
        int end = ipS2I(endIp);
        while(start <= end) {
            int firstNonZero = start & -start;
            int maxMask = 32 - log2(firstNonZero);
            maxMask = Math.max(maxMask, 32 - log2(end - start + 1));
            res.add(ipI2S(start) + "/" + maxMask);
            start += 1 << (32 - maxMask);
        }
    }catch(Exception e) {
        return res;
    }

    return res;
}

public static int ipS2I(String ip) throws Exception {
    String[] sa = ip.split("\\.");
    if (sa.length != 4) {
        throw new Exception("Bad ip address");
    }
    int res = 0;
    for (int i = 0; i < 4; i++) {
        int t = Integer.valueOf(sa[i]);
        if (t < 0 || t > 255) {
            throw new Exception("Bad ip address");
        }
        res += t << ((3-i) << 3);
    }
    return res;
}

public static String ipI2S(int ip) {
    StringBuilder sb = new StringBuilder();
    sb.append((ip>>24) & 0xFF).append(".").append((ip>>16)&0xFF).append(".").append((ip>>8) & 0xFF).append(".").append(ip & 0xFF);
    return sb.toString();
}

Answer 7

我们研究了很多，最后是这样实现的，添加下面的类后使用RouteNotation.fromRange(...)方法得到List<RouteNotation>：

package my.vpn.utils;

import java.util.ArrayList;
import java.util.List;

public class RouteNotation {
    private int ipAddress;
    private int prefix;

    private RouteNotation(long ipAddress, int prefix) {
        this.ipAddress = (int) ipAddress;
        this.prefix = prefix;
    }

    public static RouteNotation fromIPv4(long ipAddress, int prefix) {
        return new RouteNotation(ipAddress, prefix);
    }

    public long getIP() {
        return this.ipAddress;
    }

    public void setIP(long v) {
        this.ipAddress = (int) v;
    }

    public int getPrefix() {
        return this.prefix;
    }

    public void setPrefix(int bitCount) {
        this.prefix = bitCount;
    }

    public long getFirst() {
        return this.ipAddress & getMaskFromPrefix(this.prefix);
    }

    public long getLast() {
        return Notation.getNext(this.ipAddress, this.prefix) - 1;
    }

    /**
     * Generates Notation list from given range.
     * <br>
     * Each IP range can result to multiple notations.
     *
     * @param first First IP included in returned notations.
     * @param last  Last IP included in notations.
     * @return Generated routing range notations.
     */
    public static List<RouteNotation> fromRange(long first, long last) {
        List<RouteNotation> result = new ArrayList<>();
        while (last >= first) {
            // Find prefix required to mask first IP in range.
            byte prefix = getPrefixFromRange(first, last);
            // At last, push current range to result list.
            result.add(RouteNotation.fromIPv4(first, prefix));
            // Skip current Notation range.
            first = getNext(first, prefix);
        }
        return result;
    }

    public static int getMaskFromPrefix(int bitCount) {
        return (int) (0xFFFFFFFF00000000L >> bitCount);
    }

    /**
     * Skips given Notation range and get first of next notation.
     * <p>
     * Used to get next IP right out of prefix range.
     */
    public static long getNext(long ip, int prefix) {
        // Forced to use left-shift or "getUnsigned(ip)", cause else Java would
        // cast result of "Math.pow(2, 32 - prefix)" method to integer
        // (and prevents any possible integer-overflow which we require).
        return (ip & getMaskFromPrefix(prefix)) + (1L << (32 - prefix));
    }

    public static byte getPrefixFromRange(long first, long possibleLast) {
        // Find max prefix required for distance (anything less is invalid).
        long distance = (possibleLast - first) + 1;
        double bitsRequiredForDistance = Math.log(distance) / Math.log(2);
        // Get max constant bit count.
        byte prefix = (byte) (32 - Math.ceil(bitsRequiredForDistance));
        // Validate and increase limit (more prefix means less range).
        while (prefix < 32) {
            // Limit difference to last IP in range (maximum).
            long max = RouteNotation.getNext(first, prefix) - 1;
            if (max > possibleLast) {
                ++prefix;
            }
            // Never allow IP less than first in range (minimum).
            else if ((first & getMaskFromPrefix(prefix)) < first) {
                ++prefix;
            } else {
                break;
            }
        }
        return prefix;
    }
}

请注意，这是一个最低需求副本（没有我们的单元测试和许多辅助方法），但您可能需要：

/**
 * @return Zero on any error, else IP v4 integer.
 */
public static long ipv4_cast(InetAddress address) {
    long result = 0;
    if (address != null) {
        byte data[] = address.getAddress();
        if (data.length <= 4) {
            for (byte b : data) {
                result <<= 8;
                result |= b & 0xFF;
            }
        }
    }
    return result;
}