Java中的.Net Guid.ToByteArray() 等价于什么

Question

如何将 Guid.ToByteArray() 的以下 .net 用法转换为 Java？

var g= Guid.Parse("9836f2b9-ba8c-42a6-b884-2e9eed9fb95a");
var ga = g.ToByteArray();

.Net Byte 数组返回

ga= [185,242,54,152,140,186,166,66,184,132,46,158,237,159,185,90]

Java 中的尝试（与 .Net 数组不匹配）

UUID uuid = UUID.fromString("9836f2b9-ba8c-42a6-b884-2e9eed9fb95a");
ByteBuffer bb = ByteBuffer.wrap(new byte[16]);
bb.putLong(uuid.getMostSignificantBits());
bb.putLong(uuid.getLeastSignificantBits());
byte[] ga= bb.array();

返回的 Java 字节数组

ga= [-72,-124,46,-98,-19,-97,-71,90,-104,54,-14,-71,-70,-116,66,-90]

更新了Guid to Base64 in Java的解决方案

UUID uuid = UUID.fromString("9836f2b9-ba8c-42a6-b884-2e9eed9fb95a");
ByteBuffer bb = ByteBuffer.wrap(new byte[16]);
bb.putLong(uuid.getMostSignificantBits());
bb.putLong(uuid.getLeastSignificantBits());
byte[] uuid_bytes = bb.array();
byte[] guid_bytes = Arrays.copyOf(uuid_bytes,uuid_bytes.length);
guid_bytes[0] = uuid_bytes[3];
guid_bytes[1] = uuid_bytes[2];
guid_bytes[2] = uuid_bytes[1];
guid_bytes[3] = uuid_bytes[0];
guid_bytes[4] = uuid_bytes[5];
guid_bytes[5] = uuid_bytes[4];
guid_bytes[6] = uuid_bytes[7];
guid_bytes[7] = uuid_bytes[6];
byte[] ga= guid_bytes;      

Answer 1

问题的替代解决方案 -

public static byte[] getByteArrayFromUUID(UUID uuid) {
    ByteBuffer mostSignificantBitsByteBuffer = ByteBuffer.allocate(Long.BYTES)
            .putLong(uuid.getMostSignificantBits());
    return ByteBuffer.allocate(Long.BYTES * 2)
            .order(ByteOrder.LITTLE_ENDIAN)
            .putShort(mostSignificantBitsByteBuffer.getShort(2))
            .putShort(mostSignificantBitsByteBuffer.getShort(0))
            .putShort(mostSignificantBitsByteBuffer.getShort(4))
            .putShort(mostSignificantBitsByteBuffer.getShort(6))
            .order(ByteOrder.BIG_ENDIAN)
            .putLong(uuid.getLeastSignificantBits())
            .array();
}

Answer 2

Java 中近似等效的类是java.util.UUID。但是，正如您所注意到的，两者并没有给出相同的字节数组。但是，如果您执行以下操作并查看 Java 给出的数组与 .NET 给出的数组：

import java.nio.ByteBuffer;
import java.util.Arrays;
import java.util.UUID;

public class Main {

  // expected from your question
  private static final int[] EXPECTED_BYTES = {
    185, 242, 54, 152, 140, 186, 166, 66, 184, 132, 46, 158, 237, 159, 185, 90
  };

  public static void main(String[] args) {
    UUID uuid = UUID.fromString("9836f2b9-ba8c-42a6-b884-2e9eed9fb95a");

    byte[] array = toByteArray(uuid);

    System.out.println("EXPECTED: " + Arrays.toString(EXPECTED_BYTES));
    System.out.println("ACTUAL  : " + Arrays.toString(toUnsignedInts(array)));
  }

  private static byte[] toByteArray(UUID uuid) {
    return ByteBuffer.allocate(16)
        .putLong(uuid.getMostSignificantBits())
        .putLong(uuid.getLeastSignificantBits())
        .array();
  }

  // for visual purposes only
  private static int[] toUnsignedInts(byte[] array) {
    int[] result = new int[array.length];
    for (int i = 0; i < array.length; i++) {
      result[i] = Byte.toUnsignedInt(array[i]);
    }
    return result;
  }
}

还有输出：

EXPECTED: [185, 242, 54, 152, 140, 186, 166, 66, 184, 132, 46, 158, 237, 159, 185, 90]
ACTUAL  : [152, 54, 242, 185, 186, 140, 66, 166, 184, 132, 46, 158, 237, 159, 185, 90]

您会看到数组几乎相等，只是某些字节的顺序不匹配。最后八个字节（即最低有效位）都匹配，但前四个字节反转，接下来的两个字节反转，接下来的两个字节也是如此。要直观地看到它：

EXPECTED: [185, 242, 54, 152, 140, 186, 166, 66, 184, 132, 46, 158, 237, 159, 185, 90]
ACTUAL  : [152, 54, 242, 185, 186, 140, 66, 166, 184, 132, 46, 158, 237, 159, 185, 90]
           |---------------|  |------|  |-----|

我不足以解释为什么存在这种差异，但this comment 在回答问题you linked to 时说：

另请参阅[Universally unique identifier - Wikipedia]“许多系统完全以大端格式编码 UUID。” “其他系统，特别是微软在其 COM/OLE 库中编组 UUID，使用混合端格式，其中 UUID 的前三个组件是小端，后两个是大端。” – Denis 2019 年 12 月 20 日 13:06

评论所针对的答案为您的问题提供了解决方案，您已将其包含在您的问题中。该解决方案只是交换字节以获得所需的效果。这是另一个不涉及创建副本数组的解决方案：

private static byte[] toByteArray(UUID uuid) {
  long mostSigBits = uuid.getMostSignificantBits();
  return ByteBuffer.allocate(16)
      .order(ByteOrder.LITTLE_ENDIAN)
      .putInt((int) (mostSigBits >> 32))
      .putShort((short) (((int) mostSigBits) >> 16))
      .putShort((short) mostSigBits)
      .order(ByteOrder.BIG_ENDIAN)
      .putLong(uuid.getLeastSignificantBits())
      .array();
}

^{注意：我对位移不是很满意，所以可能有一种更简洁的方法来完成上述我想不到的。}

它给出以下输出：

EXPECTED: [185, 242, 54, 152, 140, 186, 166, 66, 184, 132, 46, 158, 237, 159, 185, 90]
ACTUAL  : [185, 242, 54, 152, 140, 186, 166, 66, 184, 132, 46, 158, 237, 159, 185, 90]

---

警告：不幸的是，我不确定您是否可以依靠任一解决方法在 100% 的时间内提供正确的字节。

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