python struct unpacking for a checksum

Question

I packed a struct using this line

# type(8) code(8) checksum(16) process id(16) sequence(16)
packet = struct.pack("bbHHh", ICMP_ECHO_REQUEST, 0, packet_checksum, pid, sequence)

And now im using a checksum calculation function which unpacks the packet in a way that the 16 bits of where the checksum should be (packet_checksum) is removed. But i don't know how it does that, and what does the sum function actually do? add up the different elements?:

n = len(pkt)
two_bytes = struct.unpack("%sH" % (n/2), pkt)
chksum = sum(two_bytes)

the code works well i'm just wondering how it works.

I'm puzzled at the use of n = len(pkt); ..."%sH" % (n/2) because, well, H is 2 bytes, so the length in bytes is taken, then divided by 2, then multiplied by the length of H, which is 2 again... n = (n/2) * 2 So this part seems pointless. — zxq9

Martin Evans Martin Evans · Accepted Answer · 2015-09-03T13:47:56

You seem to be trying to create an ICMP packet and to then calculate the checksum for it. For this you should carry out your initial pack with 0 for the checksum rather than trying to remove it. You could then extract each WORD as follows:

for x in xrange(0, len(packet), 2):
    print struct.unpack("H", packet[x:x+2])[0]

This assumes your packet is an even number of bytes, so if the length is odd, you could possibly temporarily append an extra 0 for the calculation.

Note, the ICMP checksum is not a simple case of adding these together, it requires a one-compliment sum. See the following answer for further information.

python struct unpacking for a checksum

1 Answers