Java中的函數getLeastSignificantBits（）＆getMostSignificantBits在Python中？

Question

有人可以幫助我形成一個簡單的函數，以提取Python中最不顯着的位數&？

防爆代碼Java中：

UUID u = UUID.fromString('a316b044-0157-1000-efe6-40fc5d2f0036'); 
long leastSignificantBits = u.getLeastSignificantBits(); 

private UUID(byte[] data) { 
    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); 
    this.mostSigBits = msb; 
    this.leastSigBits = lsb; 
} 

- >輸出值：-1160168401362026442

Answer 1

efe640fc5d2f0036十進制是17286575672347525174.。減去0x10000000000000000從中&否定：你拿到-1160168401362026442

int("efe640fc5d2f0036",16)-0x10000000000000000 -> -1160168401362026442 

請注意，這只是猜測，但似乎與您提供的唯一測試用例一起工作（f幸運的是它是負面的）。稱之爲逆向工程。

取2個最後的十六進制值（破折號分隔）並加入它們。我想，存儲裝置，當第一個數字是上述7就變成負的，所以具有較高的2-功率否定它，如果是這種情況：

def getLeastSignificantBits(s): 
    hv = "".join(s.split("-")[-2:]) 
    v = int(hv,16) 
    if int(hv[0],16)>7: 
     # negative 
     v = v-0x10000000000000000 
    return v 

print(getLeastSignificantBits('a316b044-0157-1000-efe6-40fc5d2f0036')) 

結果：

-1160168401362026442 

編輯：提供一種方法這需要整個字符串，並返回LSB & MSB夫婦

def getLeastMostSignificantBits(s): 
    sp=s.split("-") 
    lsb_s = "".join(sp[-2:]) 
    lsb = int(lsb_s,16) 
    if int(lsb_s[0],16)>7: 
     # negative 
     lsb = lsb-0x10000000000000000 

    msb_s = "".join(sp[:3]) 
    msb = int(msb_s,16) 
    if int(msb_s[0],16)>7: 
     # negative 
     msb = msb-0x10000000000000000 

    return lsb,msb 

print(getLeastMostSignificantBits('a316b044-0157-1000-efe6-40fc5d2f0036')) 

結果：

(-1160168401362026442, -6694969989912915968)