Question

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int to binary python

I have this sequence number I need to increment, now it’s fixed and working fine.
But, I just found out that actually the sequence number is not simply a short, consider this scheme:

     12 bits           4 bits
 Sequence number   Fragment number

So if I send a sequence number with the value 258 (\x02\x01), it will actually be;

Fragment number :2
Sequence number :16

I’m wondering how i can have that 258 to be the actual sequence number, and not splitted this way.
Is there any simple way to achieve this with struct ?

UPDATE:

What i meant in my above question was:

I’m building a network packet generator that goes like this:

class PacketHeader(Packet):
fields = OrderedDict([
    ("head", "\x00"),
    ("headpad", "\x00"),
    ("headlen", "\x1a\x00" ),
    ("presentflag", "\x2e"),
    ("timestamp", "\x43\x10\x58\x16\xa1\x01\x00\x00"),
    ("flag", "\x03"),
    ("Sequencenum", "\x04\x00"), 
###12 bits of this 'short' is Seq num, 4 bits left are if this is a fragment or not(if it is, its value will be =>1, and will then represent the frag number.)

])

Later, I call my class and send the packet like this:

send(PacketHeader(Sequencenum=257))
send(PacketHeader(Sequencenum=258))
send(PacketHeader(Sequencenum=259))

I need to convert to a short the sequence number (in this case lets say 257), and take care of the 4 bits that represent the fragment number in that 16 bit short. I would prefer to always have fragment number set to 0, but have Sequencenum incremented correctly.

Thanks in advance !

Answer 1

Something along the line:

number_to_send = (sequence_number << 4) + fragment_number
bytes_to_send = struct.pack("!H", number_to_send)

And back:

sequence_number = struct.unpack("!H", bytes_to_send)[0] >> 4