17

我应该如何执行从 IPv6 到 long 的转换,反之亦然?

到目前为止,我有:

    public static long IPToLong(String addr) {
            String[] addrArray = addr.split("\\.");
            long num = 0;
            for (int i = 0; i < addrArray.length; i++) {
                    int power = 3 - i;

                    num += ((Integer.parseInt(addrArray[i], 16) % 256 * Math.pow(256, power)));
            }
            return num;
    }

    public static String longToIP(long ip) {
            return ((ip >> 24) & 0xFF) + "."
                    + ((ip >> 16) & 0xFF) + "."
                    + ((ip >> 8) & 0xFF) + "."
                    + (ip & 0xFF);

    }

这是正确的解决方案还是我错过了什么?

(如果解决方案对 ipv4 和 ipv6 都有效,那就完美了)

4

4 回答 4

23

您还可以使用 java.net.InetAddress
它适用于 ipv4 和 ipv6(所有格式)

public static BigInteger ipToBigInteger(String addr) {
    InetAddress a = InetAddress.getByName(addr)
    byte[] bytes = a.getAddress()
    return new BigInteger(1, bytes)
}
于 2016-01-19T15:57:53.433 回答
10

IPv6 地址是一个 128 位数字,如此所述。Java 中的 long 以 64 位表示,因此您需要另一种结构,例如 BigDecimal 或两个 long(具有两个 long 数组的容器或简单的两个 long 数组)来存储 IPv6 地址。

下面是一个例子(只是为了给你一个想法):

public class Asd {

    public static long[] IPToLong(String addr) {
        String[] addrArray = addr.split(":");//a IPv6 adress is of form 2607:f0d0:1002:0051:0000:0000:0000:0004
        long[] num = new long[addrArray.length];
        
        for (int i=0; i<addrArray.length; i++) {
            num[i] = Long.parseLong(addrArray[i], 16);
        }
        long long1 = num[0];
        for (int i=1;i<4;i++) {
            long1 = (long1<<16) + num[i];
        }
        long long2 = num[4];
        for (int i=5;i<8;i++) {
            long2 = (long2<<16) + num[i];
        }
        
        long[] longs = {long2, long1};
        return longs;
    }
    
    
    public static String longToIP(long[] ip) {
        String ipString = "";
        for (long crtLong : ip) {//for every long: it should be two of them

            for (int i=0; i<4; i++) {//we display in total 4 parts for every long
                ipString = Long.toHexString(crtLong & 0xFFFF) + ":" + ipString;
                crtLong = crtLong >> 16;
            }
        }
        return ipString;
    
    }
    
    static public void main(String[] args) {
        String ipString = "2607:f0d0:1002:0051:0000:0000:0000:0004";
        long[] asd = IPToLong(ipString);
        
        System.out.println(longToIP(asd));
    }
}
于 2013-07-26T08:20:25.083 回答
8

IPv6 地址不能长存储。您可以使用 BigInteger 而不是 long。

public static BigInteger ipv6ToNumber(String addr) {
    int startIndex=addr.indexOf("::");

    if(startIndex!=-1){


        String firstStr=addr.substring(0,startIndex);
        String secondStr=addr.substring(startIndex+2, addr.length());


        BigInteger first=ipv6ToNumber(firstStr);

        int x=countChar(addr, ':');

        first=first.shiftLeft(16*(7-x)).add(ipv6ToNumber(secondStr));

        return first;
    }


    String[] strArr = addr.split(":");

    BigInteger retValue = BigInteger.valueOf(0);
    for (int i=0;i<strArr.length;i++) {
        BigInteger bi=new BigInteger(strArr[i], 16);
        retValue = retValue.shiftLeft(16).add(bi);
    }
    return retValue;
}


public static String numberToIPv6(BigInteger ipNumber) {
    String ipString ="";
    BigInteger a=new BigInteger("FFFF", 16);

        for (int i=0; i<8; i++) {
            ipString=ipNumber.and(a).toString(16)+":"+ipString;

            ipNumber = ipNumber.shiftRight(16);
        }

    return ipString.substring(0, ipString.length()-1);

}

public static int countChar(String str, char reg){
    char[] ch=str.toCharArray();
    int count=0;
    for(int i=0; i<ch.length; ++i){
        if(ch[i]==reg){
            if(ch[i+1]==reg){
                ++i;
                continue;
            }
            ++count;
        }
    }
    return count;
}
于 2013-10-08T04:15:41.530 回答
1

维诺德的回答是对的。但仍有一些可以改进的地方。

首先,在“countChar”方法中,“continue”应替换为“break”。

其次,必须考虑一些边界条件。

public static BigInteger ipv6ToNumber(String addr) {
    int startIndex = addr.indexOf("::");
    if (startIndex != -1) {
        String firstStr = addr.substring(0, startIndex);
        String secondStr = addr.substring(startIndex + 2, addr.length());
        BigInteger first = new BigInteger("0");
        BigInteger second = new BigInteger("0");
        if (!firstStr.equals("")) {
            int x = countChar(addr, ':');
            first = ipv6ToNumber(firstStr).shiftLeft(16 * (7 - x));
        }
        if (!secondStr.equals("")) {
            second = ipv6ToNumber(secondStr);
        }
        first = first.add(second);
        return first;
    }

    String[] strArr = addr.split(":");
    BigInteger retValue = BigInteger.valueOf(0);
    for (int i = 0; i < strArr.length; i++) {
        BigInteger bi = new BigInteger(strArr[i], 16);
        retValue = retValue.shiftLeft(16).add(bi);
    }
    return retValue;
}

public static int countChar(String str, char reg){
    char[] ch=str.toCharArray();
    int count=0;
    for(int i=0; i<ch.length; ++i){
        if(ch[i]==reg){
            if(ch[i+1]==reg){
                ++i;
                break;
            }
            ++count;
        }
    }
    return count;
}
于 2018-11-28T03:10:50.017 回答