使用python计算字符串的HMAC时出现问题

2024-06-28 11:18:28 发布

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我正试图写一个程序来计算字符串的hmac。我可以使用hashlib库,但不能使用hmac。这是我的密码:

from hashlib import sha256

def string_to_decimal(string):
    res = ''
    for char in string:
        res += hex(ord(char))[2:]
    return int(res, 16)

def decimal_to_string(number):
    hex_number = hex(number)[2:]
    hex_number = hex_number[::-1]
    res = ''
    for i in range(0, len(hex_number), 2):
        if i + 1 < len(hex_number):
            res += chr(int(hex_number[i + 1] + hex_number[i], 16))
        else:
            res += chr(int(hex_number[i], 16))

    return res[::-1]

def calculate_ipad_opad(block_size, ipad, opad):
    res_ipad = 0
    res_opad = 0
    for i in range(block_size // 8):
        res_ipad = (res_ipad * 256) + ipad
        res_opad = (res_opad * 256) + opad
    return res_ipad, res_opad

def integer_to_bytes(number):
    res = list()
    while number > 0:
        res.append(number % 256)
        number //= 256
    
    return bytes(res[::-1])

block_size = 512

msg = 'The quick brown fox jumps over the lazy dog'
key = 'key'

ipad = 54
opad = 92

block_ipad, block_opad = calculate_ipad_opad(block_size, ipad, opad)

key_decimal = string_to_decimal(key)
si = key_decimal ^ block_ipad
so = key_decimal ^ block_opad

a = sha256(integer_to_bytes(so) + sha256(integer_to_bytes(si) + msg.encode()).digest())

print(a.digest())

我知道密钥长度不超过块大小。我用Wikipedia来写代码。但它不能正常工作。你能帮我吗

编辑: 我希望代码的输出是f7bc83f430538424b13298e6aa6fb143ef4d59a14946175997479dbc2d1a3cd8。但是输出是3d6243123b984bcc17cb96eb61c2b47d27545c3a9119b623be7932e846bf0643


Tags: tokeynumbersizestringreturnbytesdef
1条回答
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1楼 · 发布于 2024-06-28 11:18:28

键的右侧必须填充0x00值,直到摘要的块大小(如果键大小小于块大小,这里就是这种情况)。您可以通过在return之前添加string_to_decimal()来实现这一点:

res = res.ljust(64 * 2, "0")

通过此更改,提供了预期的结果

对于实现,最好使用更详细的算法规范,例如FIPS PUB 198-1,而不是维基百科中的高度缩写描述。此外,如果申请官方test vectors,你是安全的


实际上不需要转换为十进制数,也就是说,可以直接使用类似字节的对象,这大大简化了实现:

from hashlib import sha256

msg = 'The quick brown fox jumps over the lazy dog'
key = 'key'

block_size_bytes = 64

block_opad_bytes = b'\x5c' * block_size_bytes
block_ipad_bytes = b'\x36' * block_size_bytes
key_bytes = key.encode().ljust(block_size_bytes, b'\0')

si_bytes = bytes([a ^ b for (a, b) in zip(block_ipad_bytes, key_bytes)]) 
so_bytes = bytes([a ^ b for (a, b) in zip(block_opad_bytes, key_bytes)]) 

a = sha256(so_bytes + sha256(si_bytes + msg.encode()).digest())

print(a.digest().hex()) # f7bc83f430538424b13298e6aa6fb143ef4d59a14946175997479dbc2d1a3cd8

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