這個Java UUID5實現沒有通過單元測試

Question

我找不到Java的自包含UUID5實現，所以我試着在下面滾動這個解決方案。它通過了我的一些單元測試，但是失敗了。這有什麼明顯的錯誤？

public static UUID UUID5(UUID namespace, String name) { 
    MessageDigest md; 
    try { 
     md = MessageDigest.getInstance("SHA-1"); 
    } catch (NoSuchAlgorithmException nsae) { 
     throw new InternalError("SHA-1 not supported"); 
    } 

    byte[] namespaceBytes = ByteBuffer.allocate(16).putLong(namespace.getMostSignificantBits()).putLong(namespace.getLeastSignificantBits()).array(); 
    byte[] nameBytes; 
    try { 
     nameBytes = name.getBytes("UTF-8"); 
    } catch (UnsupportedEncodingException e) { 
     throw new InternalError("UTF-8 encoding not supported"); 
    } 

    byte[] toHashify = new byte[namespaceBytes.length + nameBytes.length]; 
    System.arraycopy(namespaceBytes, 0, toHashify, 0, namespaceBytes.length); 
    System.arraycopy(nameBytes, 0, toHashify, namespaceBytes.length, nameBytes.length); 

    byte[] data = md.digest(toHashify); 
    data = Arrays.copyOfRange(data, 0, 16); // Wikipedia says "Note that the 160 bit SHA-1 hash is truncated to 128 bits to make the length work out." 

    data[6] &= 0x0f; /* clear version  */ 
    data[6] |= 0x50; /* set to version 5  TODO is this the correct way to do it  */ 
    data[8] &= 0x3f; /* clear variant  */ 
    data[8] |= 0x80; /* set to IETF variant */ 

    long msb = 0; 
    long lsb = 0; 
    assert data.length == 16 : "data must be 16 bytes in length"; 
    for (int i=0; i<8; i++) 
     {msb = (msb << 8) | (data[i] & 0xff);} 
    for (int i=8; i<16; i++) 
     {lsb = (lsb << 8) | (data[i] & 0xff);} 

    return new UUID(msb, lsb); 
} 

Answer 1

使用Apache Commons Id，更具體地爲UUID.nameUUIDFromString。

UUID uuid5 = UUID.nameUUIDFromString("www.ford.com", 
     UUID.fromString("078d4e79-244f-440e-844d-9454eadfd411"), 
     UUID.SHA1_ENCODING); 

UUID.SHA1_ENCODING將產生第5點的UUID。