Byte oriented data types in python

"Martin v. Löwis" martin at v.loewis.de
Sun Jan 25 23:28:55 CET 2009


>   dtype = ord(rawdata[0])
>   dcount = struct.unpack("!H",rawdata[1:3])
>   if dtype == 1:
>      fmtstr = "!" + "H"*dcount
>   elif dtype == 2:
>      fmtstr = "!" + "f"*dcount
>   rlen = struct.calcsize(fmtstr)
>   
>   data = struct.unpack(fmtstr,rawdata[3:3+rlen])
>   
>   leftover = rawdata[3+rlen:]

Unfortunately, that does not work in the example. We have
a message type (an integer), and a variable-length string.
So how do you compute the struct format for that?

>> Sure. You would normally have a struct such as
>>
>> struct TLV{
>>   char type;
>>   char length;
>>   char *data;
>> };
>>
>> However, the in-memory representation of that struct is *not*
>> meant to be sent over the wire. In particular, the character
>> pointer has no meaning outside the address space, and is thus
>> not to be sent.
> 
> Well if it's not representing the layout of the data we're
> trying to deal with, then it's irrelevent.  We are talking
> about how convert python objects to/from data in the
> 'on-the-wire' format, right?

Right: ON-THE-WIRE, not IN MEMORY. In memory, there is a
pointer. On the wire, there are no pointers.

> Like this?
> 
>>>> def encode(type,length,value):
> ...  return chr(type)+chr(length)+value
> ... 
>>>> print encode('float', 1, 3.14159)
> Traceback (most recent call last):
>   File "<stdin>", line 1, in <module>
>   File "<stdin>", line 2, in encode
> TypeError: an integer is required

No:

py> CONNECT_REQUEST=17
py> payload="call me"
py> encode(CONNECT_REQUEST, len(payload), payload)
'\x11\x07call me'

Regards,
Martin




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