Question

我需要使用XOR计算十六进制串行字符串的校验和。对于我的（有限的）知识，这必须使用按位运算符^来执行。此外，数据必须转换为二进制整数形式。下面是我的基本代码 - 但它计算的校验和是1000831.它应该是01001110或47hex。我认为错误可能是由于错过了前导零。我试图添加前导零的所有格式都将二进制整数转换回字符串。我很感激任何建议。

    word = ('010900004f')

    #divide word into 5 separate bytes
    wd1 = word[0:2] 
    wd2 = word[2:4]
    wd3 = word[4:6]
    wd4 = word[6:8]
    wd5 = word[8:10]

    #this converts a hex string to a binary string
    wd1bs = bin(int(wd1, 16))[2:] 
    wd2bs = bin(int(wd2, 16))[2:]
    wd3bs = bin(int(wd3, 16))[2:]
    wd4bs = bin(int(wd4, 16))[2:]

    #this converts binary string to binary integer
    wd1i = int(wd1bs)
    wd2i = int(wd2bs)
    wd3i = int(wd3bs)
    wd4i = int(wd4bs)
    wd5i = int(wd5bs)

    #now that I have binary integers, I can use the XOR bitwise operator to cal cksum
    checksum = (wd1i ^ wd2i ^ wd3i ^ wd4i ^ wd5i)

    #I should get 47 hex as the checksum
    print (checksum, type(checksum))

Answer 1

为什么要使用所有这些转换和昂贵的字符串函数？

（我会回答你XY-Problem的X部分，而不是Y部分。）

def checksum (s):
    v = int (s, 16)
    checksum = 0
    while v:
        checksum ^= v & 0xff
        v >>= 8
    return checksum

cs = checksum ('010900004f')
print (cs, bin (cs), hex (cs) )

结果是预期的0x47。顺便说一句0x47是0b1000111，而不是如0b1001110所述。

Answer 2

只需像这样修改。

之前：

wd1i = int(wd1bs)
wd2i = int(wd2bs)
wd3i = int(wd3bs)
wd4i = int(wd4bs)
wd5i = int(wd5bs)

后：

wd1i = int(wd1bs, 2)
wd2i = int(wd2bs, 2)
wd3i = int(wd3bs, 2)
wd4i = int(wd4bs, 2)
wd5i = int(wd5bs, 2)

为什么您的代码不起作用？

因为您误解了int(wd1bs)行为。请参阅文档here。所以Python int函数expect wd1bs默认为10 base。但是你希望int函数将其参数视为2个基数。所以你需要写成int(wd1bs, 2)

或者您也可以像这样重写整个代码。因此，在这种情况下您不需要使用bin函数。这段代码与@Hyperboreus答案基本相同。：）

w = int('010900004f', 16)
w1 = (0xff00000000 & w) >> 4*8
w2 = (0x00ff000000 & w) >> 3*8
w3 = (0x0000ff0000 & w) >> 2*8
w4 = (0x000000ff00 & w) >> 1*8
w5 = (0x00000000ff & w)

checksum = w1 ^ w2 ^ w3 ^ w4 ^ w5

print hex(checksum)
#'0x47'

这个更短。

import binascii
word = '010900004f'
print hex(reduce(lambda a, b: a ^ b, (ord(i) for i in binascii.unhexlify(word))))
#0x47

Answer 3

s = '010900004f'
b = int(s, 16)
print reduce(lambda x, y: x ^ y, ((b>> 8*i)&0xff for i in range(0, len(s)/2)), 0)

如何在二进制整数（python）中保持前导零？

3 个答案: