将IPv6转换为长到IPv6的转换

Question

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？

(如果解决方案适用于ipv4和ipv6,那将是完美的)

Answer 1

你也可以使用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)
}

Answer 2

IPv6地址是如所描述的一个128位的数目在这里.Java中的long表示为64位,因此您需要另一个结构,如BigDecimal或两个long(一个具有两个long的数组的容器或者只是一个由两个long组成的数组),以便存储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));
}

}

Answer 3

An IPv6 address can not be stored in long. You can use BigInteger instead of 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;
}