Crypto plaintext padding (SOLVED)

[David Clymer]

On Sun, 2007-05-06 at 11:39 -0400, David Clymer wrote:
> I'm using pycrypto's AES module, and am attempting to automate the
> padding of the plaintext (the size of the text to be encrypted must be a
> multiple of 16). However, my padding scheme sometimes fails to calculate
> the necessary padding correctly, and I'm not sure why.
> 
> Using the text 'foo bar', it fails, but if I do 'foo bars' it works.
> Randomized testing fails sometimes and not others.
> 
> Any pointers as to what I might be doing wrong (code attached)?

Nevermind. After I walked away from it for a bit, the mistake was
obvious. I was using the modulus of the text length by the block size
directly, rather than the block size minus that number:

--- /home/david/Desktop/test.py 2007-05-06 13:38:52.000000000 -0400
+++ test.py     2007-05-06 13:39:38.000000000 -0400
@@ -9,12 +9,12 @@
     def __init__(self, text, key_phrase):
         key = SHA256.new(key_phrase)
         self.aes = AES.new(key.digest())
-        #self.encrypted_text = self.aes.encrypt(self._pad(text))
+        self.encrypted_text = self.aes.encrypt(self._pad(text))

     def _unpad(self, txt):
         """Remove padding from the given text"""
         for x in xrange(len(txt) - self.aes.block_size,len(txt)):
-            if x == ord(txt[x]):
+            if len(txt[x:]) == ord(txt[x]):
                 if txt[x-1:] + self._gen_pad(txt[x-1:]):
                     return txt[:x]
         return txt
@@ -25,7 +25,7 @@

     def _gen_pad(self, txt):
         """Generate padding for the given plaintext"""
-        pad_len = (len(txt)) % self.aes.block_size
+        pad_len = self.aes.block_size - (len(txt) %
self.aes.block_size)
         if pad_len > 0:
             return chr(pad_len) * pad_len
         return chr(self.aes.block_size) * self.aes.block_size

-- 
gpg-key: http://www.zettazebra.com/files/key.gpg
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