compare of the CBC and CTR mode in PyCrypto AES

因为参加coursera课程cryptography I课程接触到AES算法，在涉及具体编程实现时，记录下来以便以后查找。
AES.MODE_CBC：

CBC key: 140b41b22a29beb4061bda66b6747e14
CBC Ciphertext 1:
4ca00ff4c898d61e1edbf1800618fb2828a226d160dad07883d04e008a7897ee\
2e4b7465d5290d0c0e6c6822236e1daafb94ffe0c5da05d9476be028ad7c1d81
PyCrypto 中 AES算法 密钥、原文、密文都是bit值，因此对于hex格式的数据，需要用decode('hex')解码：

from Crypto.Cipher import AES

key   = '140b41b22a29beb4061bda66b6747e14'
ctext = '4ca00ff4c898d61e1edbf1800618fb2828a226d160dad07883d04e008a7897ee2e4b7465d5290d0c0e6c6822236e1daafb94ffe0c5da05d9476be028ad7c1d81'
iv =    '4ca00ff4c898d61e1edbf1800618fb28'
cipher = AES.new(key.decode('hex') , AES.MODE_CBC , iv.decode('hex') )
text = cipher.decrypt(ctext.decode('hex'))
print text[16:] # only need msg except iv

AES.MODE_CTR：
CTR key: 36f18357be4dbd77f050515c73fcf9f2

CTR Ciphertext 2:
770b80259ec33beb2561358a9f2dc617e46218c0a53cbeca695ae45faa8952aa\
0e311bde9d4e01726d3184c34451

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

key = '36f18357be4dbd77f050515c73fcf9f2'.decode('hex') 
ct =  'e46218c0a53cbeca695ae45faa8952aa\
0e311bde9d4e01726d3184c34451'.decode('hex')
iv =  '770b80259ec33beb2561358a9f2dc617' 
ctr = Counter.new(128,initial_value = int(iv,16))

cipher = AES.new(key , AES.MODE_CTR , counter=ctr)
pt = cipher.decrypt(ct)
print pt

对于CTR模式，使用PyCrypto中的Crypto.Util.Counter进行counter值的初始化，要注意，此时解密不需要将该counter值加入解密密文，这是CTR和CBC加密原理不同造成的。当然，对于更多有关counter的内容，请访问PyCrypto的文档查找，类似的问题在Stack Overflow上也可以看到
[1]PyCrypto problem using AES+CTR
[2]Include nonce and block count in PyCrypto AES MODE_CTR