From: Rob Biedenharn <Rob@...>
Date: 2009-03-24T12:43:39+09:00
Subject: Re: Mimic AES_ENCRYPT and AES_DECRYPT functions in Ruby

On Mar 23, 2009, at 11:25 PM, Felipe Coury wrote:

> Because AES is a block-level algorithm, padding is used to encode  
> uneven
> length strings and so the result string length may be calculated using
> this formula:
>
> 16 × (trunc(string_length / 16) + 1)


Do you mean to have:

16 * (string_length + 15)/16

If the string length is 32, what do you expect the result to be? Your  
formula gives 48, (16*(trunc(32/16)+1))==(16*(2+1)), while mine gives  
32, (16*(32+15)/16)==(16*(47/16))==(16*2) [integer division].

Here's a bit of code that I've lifted out of another project:

     # Encrypt the content of the document, block by block, in a manner
     # compatible with the original Python (so we can decrypt it and
     # remain backwardly compatible)
     rijndael = Crypt::Rijndael.new(self.key, 256, 256)
     encryptedData = ""
     data << 'X'            # a marker added to cope with partial block
     blocks, bytes = data.length.divmod(32)
     unless bytes.zero?
       data << "\0" * (32 - bytes)
       blocks += 1
     end
     (0...blocks).each do |block|
       encryptedData << rijndael.encrypt_block(data[block * 32, 32])
     end

The decrypting side was Java and I don't know why the 'X' was chosen  
(seems that I recall something about there being a byte with the  
number of extra bytes of padding... or your formula might hold a clue).

Anyway, you'd have to adjust it for 128-bit/16-byte keys (and blocks).

-Rob

Rob Biedenharn		http://agileconsultingllc.com
Rob@AgileConsultingLLC.com