GUID to ByteArray | 易学教程

问题

I just wrote this code to convert a GUID into a byte array. Can anyone shoot any holes in it or suggest something better?

 public static byte[] getGuidAsByteArray(){

 UUID uuid = UUID.randomUUID();
 long longOne = uuid.getMostSignificantBits();
 long longTwo = uuid.getLeastSignificantBits();

 return new byte[] {
      (byte)(longOne >>> 56),
      (byte)(longOne >>> 48),
      (byte)(longOne >>> 40),
      (byte)(longOne >>> 32),   
      (byte)(longOne >>> 24),
      (byte)(longOne >>> 16),
      (byte)(longOne >>> 8),
      (byte) longOne,
      (byte)(longTwo >>> 56),
      (byte)(longTwo >>> 48),
      (byte)(longTwo >>> 40),
      (byte)(longTwo >>> 32),   
      (byte)(longTwo >>> 24),
      (byte)(longTwo >>> 16),
      (byte)(longTwo >>> 8),
      (byte) longTwo
       };
}

In C++, I remember being able to do this, but I guess theres no way to do it in Java with the memory management and all?:

    UUID uuid = UUID.randomUUID();

    long[] longArray = new long[2];
    longArray[0] = uuid.getMostSignificantBits();
    longArray[1] = uuid.getLeastSignificantBits();

    byte[] byteArray = (byte[])longArray;
    return byteArray;

Edit

If you want to generate a completely random UUID as bytes that does not conform to any of the official types, this will work and wastes 10 fewer bits than type 4 UUIDs generated by UUID.randomUUID():

    public static byte[] getUuidAsBytes(){
    int size = 16;
    byte[] bytes = new byte[size];
    new Random().nextBytes(bytes);
    return bytes;
}

回答1:

I would rely on built in functionality:

ByteBuffer bb = ByteBuffer.wrap(new byte[16]);
bb.putLong(uuid.getMostSignificantBits());
bb.putLong(uuid.getLeastSignificantBits());
return bb.array();

or something like,

ByteArrayOutputStream ba = new ByteArrayOutputStream(16);
DataOutputStream da = new DataOutputStream(ba);
da.writeLong(uuid.getMostSignificantBits());
da.writeLong(uuid.getLeastSignificantBits());
return ba.toByteArray();

(Note, untested code!)

回答2:

public static byte[] newUUID() {
    UUID uuid = UUID.randomUUID();
    long hi = uuid.getMostSignificantBits();
    long lo = uuid.getLeastSignificantBits();
    return ByteBuffer.allocate(16).putLong(hi).putLong(lo).array();
}

回答3:

You can check UUID from apache-commons. You may not want to use it, but check the sources to see how its getRawBytes() method is implemented:

public UUID(long mostSignificant, long leastSignificant) {
    rawBytes = Bytes.append(Bytes.toBytes(mostSignificant), Bytes.toBytes(leastSignificant));
}

回答4:

You could take a look at Apache Commons Lang3 Conversion.uuidToByteArray(...). Conversely, look at Conversion.byteArrayToUuid(...) to convert back to a UUID.

来源：https://stackoverflow.com/questions/2983065/guid-to-bytearray

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