Re: Representation clauses for base-64 encoding

["Dmitry A. Kazakov" <mailbox@dmitry-kazakov.de>]

On Wed, 28 Dec 2011 23:41:57 -0600, Randy Brukardt wrote:

> "Niklas Holsti" <niklas.holsti@tidorum.invalid> wrote in message 
> news:9m0433FjokU1@mid.individual.net...

>> True for internal data, false for inputs and outputs.
> 
> Not really. The vast majority of inputs and outputs are in text form (as in 
> the encoding example). Most of the rest are just a pure stream of bytes. And 
> the interesting thing with a stream of bytes is that you get identical 
> behavior for both bit orders -- because the bit order and byte order are 
> tied together. The problem comes when you try to use the non-native bit and 
> byte order on a machine.

No. It is true that bit order within the octets is properly handled by the
hardware, wrong is that you could ignore it once you got the octets read.
Many octet-based protocols use bit fields spanning across several octets
starting and ending anywhere in the octet. Probably you could ignore this
reordering octets of the stream into the machine-native order. If you knew
how to, because of nice middle-endian representations.

Formally speaking a stream cannot be reordered at all, because it is
unbounded. You could reorder only octets of a packet, a word etc. But not
before calculating the check sums. The real-life protocols are difficult to
crack.

-- 
Regards,
Dmitry A. Kazakov
http://www.dmitry-kazakov.de