将IPv6转换为长到IPv6的转换

时间:2013-07-26 07:45:34

标签: java ip type-conversion ipv6 long-integer

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遗漏了什么?

(如果解决方案适用于ipv4和ipv6,那将是完美的)

4 个答案:

答案 0 :(得分:16)

您也可以使用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)
}

答案 1 :(得分:9)

IPv6地址是一个128位数字,如here所述。 Java中的长整数以64位表示,因此您需要另一种结构,如BigDecimal或两个long(具有两个long的数组的容器或者只是两个long的数组),以便存储IPv6地址。

以下是一个示例(仅为您提供一个想法):

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));
}

}

答案 2 :(得分:7)

无法长时间存储IPv6地址。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;
}

答案 3 :(得分:1)

维诺德的答案是正确的。但是仍然有一些可以改进的地方。(抱歉,由于我是新来的,我无法在Vinod的回答下方添加评论。)

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

第二,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;
}