python - Python : HMAC Library vs Hashlib Produces Different Results 中的 HMAC SHA256

Question

所以我一直在查看维基百科的 HMAC 伪代码，它看起来相对简单；如果您的 key 大小已经是块大小，则伪代码可归结为 3 行:

    o_key_pad ← key xor [0x5c * blockSize]   // Outer padded key
    i_key_pad ← key xor [0x36 * blockSize]   // Inner padded key

    return hash(o_key_pad ∥ hash(i_key_pad ∥ message))

这很容易转化为 Python:

ik = bytes([0x36 ^ b for b in k])
ok = bytes([0x5c ^ b for b in k])

print(hashlib.sha256(ok + bytearray.fromhex(hashlib.sha256(ik+msg).hexdigest())).hexdigest())

但这不会产生与使用 Python 的 HMAC 库相同的结果:

p = bytes("password", encoding='utf8')
k = bytearray.fromhex("eae1f9b8c78d0e0dbaeb3bc49fea3f0be9e9dc580c0b0ba09bcf5104713fda80")

x = hmac.new(k, digestmod='sha256')
x.update(p)
print(x.hexdigest())

ik = bytes([0x36 ^ b for b in k])
ok = bytes([0x5c ^ b for b in k])
print(hashlib.sha256(ok + bytearray.fromhex(hashlib.sha256(ik+p).hexdigest())).hexdigest())

最终生产
1b96a6d3473698c3592a99d752934b875f82cdd623230abc534f92e7b70cc251
57dcbe4ada890bcc8d3cc2e6072874e0d1a0d6d3f73ceb1ced8dad4f07b56e33
为什么？

Answer 1

在自定义实现中，缺少块大小键的填充，参见 here :

Keys shorter than blockSize are padded to blockSize by padding with zeros on the right

SHA256的块大小为64字节，见here .
以下 ptyhon 代码产生预期结果:

import hmac 
import hashlib

p = bytes("password", encoding='utf8')
k = bytearray.fromhex("eae1f9b8c78d0e0dbaeb3bc49fea3f0be9e9dc580c0b0ba09bcf5104713fda80")

x = hmac.new(k, digestmod='sha256')
x.update(p)
print(x.hexdigest())

k = bytearray.fromhex("eae1f9b8c78d0e0dbaeb3bc49fea3f0be9e9dc580c0b0ba09bcf5104713fda80".ljust(128,'0')) # pad with zeros
ik = bytes([0x36 ^ b for b in k])
ok = bytes([0x5c ^ b for b in k])
print(hashlib.sha256(ok + bytearray.fromhex(hashlib.sha256(ik+p).hexdigest())).hexdigest())

输出:

1b96a6d3473698c3592a99d752934b875f82cdd623230abc534f92e7b70cc251
1b96a6d3473698c3592a99d752934b875f82cdd623230abc534f92e7b70cc251

也可以验证here .