在Python中使用paramiko生成SSH密钥对

Question

我正在尝试使用Python的paramiko模块生成SSH密钥对。关于密钥生成的信息似乎不多。我看过paramiko的文档，但还是搞不清楚哪里出了问题。我可以生成没有密码加密的私钥和公钥。但是，当我尝试加密私钥时，出现了以下错误。

ValueError: IV必须是8个字节长

我认为这个错误是来自pycrypto。我查看了paramiko.pkey和pycrypto中的相关代码，但没有找到解决办法。

这里有一个小例子。

import paramiko

def keygen(filename,passwd=None,bits=1024):
    k = paramiko.RSAKey.generate(bits)
    #This line throws the error.
    k.write_private_key_file(filename,password = 'cleverpassword')
    o = open(fil+'.pub' ,"w").write(k.get_base64())

错误追踪信息

Traceback (most recent call last):
  File "/var/mobile/Applications/149E4C21-2F92-4712-BAC6-151A171C6687/Documents/test.py", line 14, in keygen
    k.write_private_key_file(filename,password = 'cleverpassword')
  File "/var/mobile/Applications/149E4C21-2F92-4712-BAC6-151A171C6687/Pythonista.app/pylib/site-packages/paramiko/rsakey.py", line 127, in write_private_key_file
    self._write_private_key_file('RSA', filename, self._encode_key(), password)
  File "/var/mobile/Applications/149E4C21-2F92-4712-BAC6-151A171C6687/Pythonista.app/pylib/site-packages/paramiko/pkey.py", line 323, in _write_private_key_file
    self._write_private_key(tag, f, data, password)
  File "/var/mobile/Applications/149E4C21-2F92-4712-BAC6-151A171C6687/Pythonista.app/pylib/site-packages/paramiko/pkey.py", line 341, in _write_private_key
    data = cipher.new(key, mode, salt).encrypt(data)
  File "/var/mobile/Applications/149E4C21-2F92-4712-BAC6-151A171C6687/Pythonista.app/pylib/site-packages/Crypto/Cipher/DES3.py", line 114, in new
    return DES3Cipher(key, *args, **kwargs)
  File "/var/mobile/Applications/149E4C21-2F92-4712-BAC6-151A171C6687/Pythonista.app/pylib/site-packages/Crypto/Cipher/DES3.py", line 76, in __init__
    blockalgo.BlockAlgo.__init__(self, _DES3, key, *args, **kwargs)
  File "/var/mobile/Applications/149E4C21-2F92-4712-BAC6-151A171C6687/Pythonista.app/pylib/site-packages/Crypto/Cipher/blockalgo.py", line 141, in __init__
    self._cipher = factory.new(key, *args, **kwargs)
ValueError: IV must be 8 bytes long

Answer 1

问题

这看起来是 paramiko 里的一个错误。

如果你查看 pkey.py 中引发错误的那一行，它是下面这一行：

        data = cipher.new(key, mode, salt).encrypt(data)

现在我们来看它之前的几行，这些行通过先选择一个 cipher_name 来设置 mode。

        # since we only support one cipher here, use it
        cipher_name = list(self._CIPHER_TABLE.keys())[0]
        cipher = self._CIPHER_TABLE[cipher_name]['cipher']
        keysize = self._CIPHER_TABLE[cipher_name]['keysize']
        blocksize = self._CIPHER_TABLE[cipher_name]['blocksize']
        mode = self._CIPHER_TABLE[cipher_name]['mode']

这里是 _CIPHER_TABLE 的内容。

_CIPHER_TABLE = {
    'AES-128-CBC': {'cipher': AES, 'keysize': 16, 'blocksize': 16, 'mode': AES.MODE_CBC},
    'DES-EDE3-CBC': {'cipher': DES3, 'keysize': 24, 'blocksize': 8, 'mode': DES3.MODE_CBC},
}

注意一下，注释和代码之间存在矛盾。可用的加密方式有两个，而上面选择 cipher_name 的那一行假设只有一个。

根据错误信息，似乎选择的是 'DES-EDE3-CBC'。如果我们查看 DES3.py 中的注释，会看到对 IV 的要求。

  IV : byte string
    The initialization vector to use for encryption or decryption.

    It is ignored for `MODE_ECB` and `MODE_CTR`.

    For `MODE_OPENPGP`, IV must be `block_size` bytes long for encryption
    and `block_size` +2 bytes for decryption (in the latter case, it is
    actually the *encrypted* IV which was prefixed to the ciphertext).
    It is mandatory.

从 paramiko 的源代码中，我们发现没有传递 IV，因此出现了我们看到的错误。

解决方法

将 pkey.py 中的以下一行修改为硬编码 'AES-128-CBC' 加密方式。

   # cipher_name = list(self._CIPHER_TABLE.keys())[1]
   cipher_name = 'AES-128-CBC'