Pycrypto：CTR模式下的计数器增量

Question

Pycrypto：CTR模式下的计数器增量

4

我仍然无法使这个工作。我的问题是如何使解密行起作用。以下是我的代码：

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' 参数必须是可调用的对象

我就是想不明白pycrypto希望我如何组织new方法的第三个参数。

有人能帮忙吗？谢谢！

编辑按照下面的建议实施后的新代码。仍然卡住了！

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

编辑：仍然无法让这个工作起来。非常沮丧，完全没有头绪。以下是最新的代码:（请注意，我的输入字符串是32位十六进制字符串，必须以两位数字对进行解释以转换为长整型。）

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计数器函数返回的字符串长度不是16

- AndroidDev

1个回答

网页内容由stack overflow 提供, 点击上面的

可以查看英文原文，
原文链接

- atomicinf · Accepted Answer

在Python中，将函数视为对象是完全有效的。同样，将任何定义了__call__(self, ...)的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)

然而，PyCrypto为你提供了一个计数器方法，它应该比纯Python更快：

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

ctr现在是一个有状态的函数（同时也是可调用对象），每次调用时会增加并返回其内部状态。然后您可以执行

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

与以前一样。

以下是使用Crypto.Cipher.AES在CTR模式下带有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)