MD5的实现Python改进代码
时间: 2023-11-05 16:19:55 浏览: 35
以下是一个改进版的Python MD5实现代码,它支持任意长度的输入字符串,并使用内置的Python `struct`库来处理二进制数据:
```python
import hashlib
import struct
def md5(data):
# initialize hash values
a0 = 0x67452301
b0 = 0xEFCDAB89
c0 = 0x98BADCFE
d0 = 0x10325476
# initialize constants
s = [7, 12, 17, 22] * 4 + [5, 9, 14, 20] * 4 + [4, 11, 16, 23] * 4 + [6, 10, 15, 21] * 4
K = [int(2**32 * abs(math.sin(i+1))) & 0xFFFFFFFF for i in range(64)]
# preprocess input data
bit_len = len(data) * 8
data += b'\x80'
while (len(data) * 8 + 64) % 512 != 0:
data += b'\x00'
data += struct.pack('<Q', bit_len)
# process input data in 512-bit chunks
for i in range(0, len(data), 64):
chunk = data[i:i+64]
words = list(struct.unpack('<16I', chunk))
a, b, c, d = a0, b0, c0, d0
for j in range(64):
if j < 16:
f = (b & c) | ((~b) & d)
g = j
elif j < 32:
f = (d & b) | ((~d) & c)
g = (5*j + 1) % 16
elif j < 48:
f = b ^ c ^ d
g = (3*j + 5) % 16
else:
f = c ^ (b | (~d))
g = (7*j) % 16
f = f & 0xFFFFFFFF
temp = d
d = c
c = b
b = b + ((a + f + K[j] + words[g]) << s[j])
b = b & 0xFFFFFFFF
a = temp
a0 = (a0 + a) & 0xFFFFFFFF
b0 = (b0 + b) & 0xFFFFFFFF
c0 = (c0 + c) & 0xFFFFFFFF
d0 = (d0 + d) & 0xFFFFFFFF
# concatenate hash values and return as hexadecimal string
return '{:08x}{:08x}{:08x}{:08x}'.format(a0, b0, c0, d0)
```
这个改进版的MD5实现代码使用了更加高效的算法来处理输入数据,并且能够处理任意长度的输入字符串。为了实现这些功能,我们用 Python 的 `struct` 模块来处理二进制数据。最终,我们将散列值连接在一起并以十六进制字符串形式返回。