Question

我正在尝试将此代码（from David Clunie）转换为C＃，以便在我的程序中从UUID（或GUID）创建OID

public static String createOIDFromUUIDCanonicalHexString(String hexString) throws IllegalArgumentException {
        UUID uuid = UUID.fromString(hexString);
        long leastSignificantBits = uuid.getLeastSignificantBits();
        long mostSignificantBits  = uuid.getMostSignificantBits();
        BigInteger decimalValue = makeBigIntegerFromUnsignedLong(mostSignificantBits);
        decimalValue = decimalValue.shiftLeft(64);
        BigInteger bigValueOfLeastSignificantBits = makeBigIntegerFromUnsignedLong(leastSignificantBits);
        decimalValue = decimalValue.or(bigValueOfLeastSignificantBits);   // not add() ... do not want to introduce question of signedness of long
        return OID_PREFIX+"."+decimalValue.toString();

我不明白为什么要从UUID的各个部分制作longs（leastSignificantBits，mostSignificantBits），然后从它们制作bigint - 为什么不直接制作BigInt呢？（因为无论如何他都会移动最重要的数字）。

任何人都可以告诉我为什么这样写的方式是什么？（免责声明：我没有尝试运行java代码，我只是想在C＃中实现它）

[编辑]

原来有几个问题 - 正如Kevin Coulombe指出的那样，一个很大的问题是微软存储GUID的字节顺序.java代码似乎从明显（从左到右）顺序得到字节，但在两个显然（也感谢Kevin）因为没有简单的方法来获取Java中的整个字节数组。

以下是C＃代码，转发和（部分）反向：

    class Program
{
    const string OidPrefix = "2.25.";

    static void Main(string[] args)
    {
        Guid guid = new Guid("000000FF-0000-0000-0000-000000000000");
        //Guid guid = new Guid("f81d4fae-7dec-11d0-a765-00a0c91e6bf6");
        Console.WriteLine("Original guid: " + guid.ToString());
        byte[] guidBytes = StringToByteArray(guid.ToString().Replace("-", ""));
        BigInteger decimalValue = new BigInteger(guidBytes);
        Console.WriteLine("The OID is " + OidPrefix + decimalValue.ToString());
        string hexGuid = decimalValue.ToHexString().PadLeft(32, '0');//padded for later use
        Console.WriteLine("The hex value of the big int is " + hexGuid);
        Guid testGuid = new Guid(hexGuid);
        Console.WriteLine("This guid should match the orginal one: " + testGuid);
        Console.ReadKey();
    }

    public static byte[] StringToByteArray(String hex)
    {
        int NumberChars = hex.Length;
        byte[] bytes = new byte[NumberChars / 2];
        for (int i = 0; i < NumberChars; i += 2)
            bytes[i / 2] = Convert.ToByte(hex.Substring(i, 2), 16);
        return bytes;
    }
}

Answer 1

我认为他们这样做是因为没有简单的方法将UUID转换为字节数组以传递给Java中的BigInteger。

请参阅：GUID to ByteArray

在C＃中，这应该是你要找的东西：

String oid = "prefix" + "." + new System.Numerics.BigInteger(
        System.Guid.NewGuid().ToByteArray()).ToString();

Answer 2

String oid_prefix = "2.25"
String hexString = "f81d4fae-7dec-11d0-a765-00a0c91e6bf6"
UUID uuid = UUID.fromString(hexString);
long leastSignificantBits = uuid.getLeastSignificantBits();
long mostSignificantBits  = uuid.getMostSignificantBits();
mostSignificantBits = mostSignificantBits & Long.MAX_VALUE;
BigInteger decimalValue = BigInteger.valueOf(mostSignificantBits);
decimalValue = decimalValue.setBit(63);
decimalValue = decimalValue.shiftLeft(64);
leastSignificantBits = leastSignificantBits & Long.MAX_VALUE;
BigInteger bigValueLeastSignificantBit = BigInteger.valueOf(leastSignificantBits);
bigValueLeastSignificantBit = bigValueLeastSignificantBit.setBit(63);
decimalValue = decimalValue.or(bigValueLeastSignificantBit);
println "oid is = "+oid_prefix+"."+decimalValue

将UUID转换为OID，将Java代码转换为C＃ - 这是如何工作的？

2 个答案: