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)
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;
}