Re: AES (Rijndael) Encryption with CryptoAPI

From: Ryan Menezes [MSFT] (ryanmen_at_online.microsoft.com)
Date: 06/25/04 

Date: Thu, 24 Jun 2004 17:42:48 -0700

Trying passing FALSE for the Final flag to CryptEncrypt() if you don't want
a whole extra block added to data which is already a multiple of block size. 

-- 
Thanks,
Ryan Menezes [MS]
This posting is provided "AS IS" with no warranties, and confers no rights.
"Pieter Philippaerts" <Pieter.nospam@mentalis.org> wrote in message
news:eBpIqkkWEHA.2576@TK2MSFTNGP10.phx.gbl...
> "Abe Simpson" <abe@simpson.com> wrote in message
> > When I encrypt plaintext that is 1 to 15 bytes long, the resultant
> > ciphertext is always 16 bytes, and a 16-byte plaintext results in a
> 32-byte
> > cipher text.
>
> This is normal behavior. The AES is a block cipher, which means it always
> operates on 16 byte blocks.
> If the input data is not an exact multiple of 16, the data will be padded
> until it has a length that is a multiple of 16.
> If the input data is an exact multiple of 16, the CryptoAPI will add a
full
> 16-byte padding block at the end of the data.
>
> > Rijndael specs provide sample results where 16-byte plaintext becomes
> > 16-byte ciphertext. How do I achieve the same result?
>
> You can simply ignore the last 16 bytes, since they are the encrypted
> padding.
>
> Regards,
> Pieter Philippaerts
>
> P.S.: in case the results of your code are not the same as the results
> you're trying to verify it with, a common mistake is that the CryptoAPI is
> little endian and virtually every other platform is big endian (Java,
.NET,
> and probably the AES reference too). So you may need to reverse the key
and
> IV.
>
>

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