python - Pycrypto:递增点击率模式

Question

仍然不能让它正常工作。我的问题是关于如何使解密线工作。这是我写的:

class IVCounter(object):
    @staticmethod
    def incrIV(self):
        temp = hex(int(self, 16)+1)[2:34]
        return array.array('B', temp.decode("hex")).tostring()


def decryptCTR(key, ciphertext):

    iv = ciphertext[:32] #extracts the first 32 characters of the ciphertext

    #convert the key into a 16 byte string
    key = array.array('B', key.decode("hex")).tostring()

    print AES.new(key, AES.MODE_CTR, counter=IVCounter.incrIV(iv)).decrypt(ciphertext)
    return

我的错误信息是:

ValueError: 'counter' parameter must be a callable object

我只是想不通 pycrypto 要我如何组织 new 的第三个参数。

有人可以帮忙吗？谢谢!

编辑实现以下建议后的新代码。还是卡住了!

class IVCounter(object):
    def __init__(self, start=1L):
        print start #outputs the number 1 (not my IV as hoped)
        self.value = long(start)

   def __call__(self):
        print self.value  #outputs 1 - need this to be my iv in long int form
        print self.value + 1L  #outputs 2
        self.value += 1L
        return somehow_convert_this_to_a_bitstring(self.value) #to be written

def decryptCTR(key, ciphertext):

    iv = ciphertext[:32] #extracts the first 32 characters of the ciphertext
    iv = int(iv, 16)

    #convert the key into a 16 byte string
    key = array.array('B', key.decode("hex")).tostring()

    ctr = IVCounter()
    Crypto.Util.Counter.new(128, initial_value = iv)

    print AES.new(key, AES.MODE_CTR, counter=ctr).decrypt(ciphertext)
    return

EDIT 仍然无法正常工作。非常沮丧，完全没有想法。这是最新的代码:(请注意，我的输入字符串是 32 位十六进制字符串，必须解释为两位数对才能转换为长整数。)

class IVCounter(object):
    def __init__(self, start=1L):
        self.value = long(start)

    def __call__(self):
        self.value += 1L
        return hex(self.value)[2:34]

def decryptCTR(key, ciphertext):
    iv = ciphertext[:32] #extracts the first 32 characters of the ciphertext
    iv = array.array('B', iv.decode("hex")).tostring()

    ciphertext = ciphertext[32:]

    #convert the key into a 16 byte string
    key = array.array('B', key.decode("hex")).tostring()

    #ctr = IVCounter(long(iv))
    ctr = Crypto.Util.Counter.new(16, iv)

    print AES.new(key, AES.MODE_CTR, counter=ctr).decrypt(ciphertext)
    return

TypeError: CTR counter function returned string not of length 16

Answer 1

在 Python 中，将函数视为对象是完全有效的。将定义 __call__(self, ...) 的任何对象视为函数也是完全有效的。

所以你想要的可能是这样的:

class IVCounter(object):
    def __init__(self, start=1L):
        self.value = long(start)
    def __call__(self):
        self.value += 1L
        return somehow_convert_this_to_a_bitstring(self.value)

ctr = IVCounter()
... make some keys and ciphertext ...
print AES.new(key, AES.MODE_CTR, counter=ctr).decrypt(ciphertext)

但是，PyCrypto 为您提供了一个计数器方法，它应该比纯 Python 快得多:

import Crypto.Util.Counter
ctr = Crypto.Util.Counter.new(NUM_COUNTER_BITS)

ctr 现在是一个有状态函数(同时也是一个可调用对象)，它会在您每次调用它时递增并返回其内部状态。然后你可以做

print AES.new(key, AES.MODE_CTR, counter=ctr).decrypt(ciphertext)

和以前一样。

这是一个在 CTR 模式下使用 Crypto.Cipher.AES 和用户指定的初始化向量的工作示例:

import Crypto.Cipher.AES
import Crypto.Util.Counter

key = "0123456789ABCDEF" # replace this with a sensible value, preferably the output of a hash
iv = "0000000000009001" # replace this with a RANDOMLY GENERATED VALUE, and send this with the ciphertext!

plaintext = "Attack at dawn" # replace with your actual plaintext

ctr = Crypto.Util.Counter.new(128, initial_value=long(iv.encode("hex"), 16))

cipher = Crypto.Cipher.AES.new(key, Crypto.Cipher.AES.MODE_CTR, counter=ctr)
print cipher.encrypt(plaintext)