How to calculate Internet checksum? - checksum

I have a question regarding how the Internet checksum is calculated. I couldn't find any good explanation from the book, so I ask it here.
Have a look at the following example.
The following two messages are sent: 10101001 and 00111001. The checksum is calculated with 1's complement. So far I understand. But how is the sum calculated? At first I thought it maybe is XOR, but it seems not to be the case.
10101001
00111001
--------
Sum 11100010
Checksum: 00011101
And then when they calculate if the message arrived OK. And once again how is the sum calculated?
10101001
00111001
00011101
--------
Sum 11111111
Complement 00000000 means that the pattern is O.K.

It uses addition, hence the name "sum". 10101001 + 00111001 = 11100010.
For example:
+------------+-----+----+----+----+---+---+---+---+--------+
| bin value | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 | result |
+------------+-----+----+----+----+---+---+---+---+--------+
| value 1 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 1 | 169 |
| value 2 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 1 | 57 |
| sum/result | 1 | 1 | 1 | 0 | 0 | 0 | 1 | 0 | 226 |
+------------+-----+----+----+----+---+---+---+---+--------+

If by internet checksum you mean TCP Checksum there's a good explanation here and even some code.
When you're calculating the checksum remember that it's not just a function of the data but also of the "pseudo header" which puts the source IP, dest IP, protocol, and length of the TCP packet into the data to be checksummed. This ties the TCP meta data to some data in the IP header.
TCP/IP Illustrated Vol 1 is a good reference for this and explains it all in detail.

The calculation of the internet checksum uses ones complement arithmetic. Consider the data being checksummed is a sequence of 8 bit integers. First you need to add them using ones complement arithmetic and take the ones complement of the result.
NOTE: When adding numbers ones complement arithmetic, a carryover from the MSB needs to be added to the result. Consider for eg., the addition of 3(0011) and 5(0101).
3'->1100
5'->1010
0110 with a carry of 1
Thus we have, 0111(1's complement representation of -8).
The checksum is the 1's complement of the result obtained int he previous step. Hence we have 1000. If no carry exists, we just complement the result obtained in the summing stage.

The UDP checksum is created on the sending side by summing all the 16-bit words in the segment, with any overflow being wrapped around and then the 1's complement is performed and the result is added to the checksum field inside the segment.
at the receiver side, all words inside the packet are added and the checksum is added upon them if the result is 1111 1111 1111 1111 then the segment is valid else the segment has an error.
exmaple:
0110 0110 0110 0000
0101 0101 0101 0101
1000 1111 0000 1100
--------------------
1 0100 1010 1100 0001 //there is an overflow so we wrap it up, means add it to the sum
the sum = 0100 1010 1100 0010
now let's take the 1's complement
checksum = 1011 0101 0011 1101
at the receiver the sum is calculated and then added to the checksum
0100 1010 1100 0010
1011 0101 0011 1101
----------------------
1111 1111 1111 1111 //clearly this should be the answer, if it isn't then there is an error
references:Computer networking a top-down approach[Ross-kurose]

Here's a complete example with a real header of an IPv4 packet.
In the following example, I use bc, printf and here strings to calculate the header checksum and verify it. Consequently, it should be easy to reproduce the results on Linux by copy-pasting the commands.
These are the twenty bytes of our example packet header:
45 00 00 34 5F 7C 40 00 40 06 [00 00] C0 A8 B2 14 C6 FC CE 19
The sender hasn't calculated the checksum yet. The two bytes in square brackets is where the checksum will go. The checksum's value is initially set to zero.
We can mentally split up this header as a sequence of ten 16-bit values: 0x4500, 0x0034, 0x5F7C, etc.
Let's see how the sender of the packet calculates the header checksum:
Add all 16-bit values to get 0x42C87: bc <<< 'obase=16;ibase=16;4500 + 0034 + 5F7C + 4000 + 4006 + 0000 + C0A8 + B214 + C6FC + CE19'
The leading digit 4 is the carry count, we add this to the rest of the number to get 0x2C8B: bc <<< 'obase=16;ibase=16;2C87 + 4'
Invert¹ 0x2C8B to get the checksum: 0xD374
Finally, insert the checksum into the header:
45 00 00 34 5F 7C 40 00 40 06 [D3 74] C0 A8 B2 14 C6 FC CE 19
Now the header is ready to be sent.
The recipient of the IPv4 packet then creates the checksum of the received header in the same way:
Add all 16-bit values to get 0x4FFFB: bc <<< 'obase=16;ibase=16;4500 + 0034 + 5F7C + 4000 + 4006 + D374 + C0A8 + B214 + C6FC + CE19'
Again, there's a carry count so we add that to the rest to get 0xFFFF: bc <<< 'obase=16;ibase=16;FFFB + 4'
If the checksum is 0xFFFF, as in our case, the IPv4 header is intact.
See the Wikipedia entry for more information.
¹Inverting the hexadecimal number means converting it to binary, flipping the bits, and converting it to hexadecimal again. You can do this online or with Bash: hex_nr=0x2C8B; hex_len=$(( ${#hex_nr} - 2 )); inverted=$(printf '%X' "$(( ~ hex_nr ))"); trunc_inverted=${inverted: -hex_len}; echo $trunc_inverted

Related

Why is there something written in the data section of an ICMPv4 echo ping request?

(My question differs from this one.)I am connected to a AP in a wireless network and I've send a simple ping request to www.google.com. When I analyze the packet in wireshark, I can see, that there are 48 bytes written in the data section of ICMP. After 8 bytes of trash values, the values are sequentially increasing from 0x10 to 0x37.Is there any particular reason, why ICMPv4 fits values instead of just using a bunch of zeroes?
The hexdump of the ICMPv4 data section:
0030 09 d9 0d 00 00 00 00 00 10 11 12 13 14 15 .Ù............
0040 16 17 18 19 1a 1b 1c 1d 1e 1f 20 21 22 23 24 25 .......... !"#$%
0050 26 27 28 29 2a 2b 2c 2d 2e 2f 30 31 32 33 34 35 &'()*+,-./012345
0060 36 37 67
After 8 bytes of trash values
First of all, these are not trash values. In some implementations of ping, the 1st 8 bytes may represent a timestamp.
As #ross-jacobs mentioned, RFC 792 describes the ICMP Echo Request/Reply Packets. For clarity, these two packets are described, in relevant part, as follows:
Echo or Echo Reply Message
0 1 2 3
0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
| Type | Code | Checksum |
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
| Identifier | Sequence Number |
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
| Data ...
+-+-+-+-+-
...
Description
The data received in the echo message must be returned in the echo
reply message.
Here you can see that the contents of the Data is not described at all; therefore an implementation is free to use whatever data it wishes, including none at all.
Now, since ping is a network test tool, one of the things it can help test is fragmentation/reassembly. Every ping implementation I'm aware of allows the user to specify the size of the payload, and if you exceed the MTU, you should see the ICMP packet fragmented/reassembled. If you examine the payload of the first fragment, you can tell where the second fragment should start just by looking at the sequence of bytes in the payload of the first fragment. If the data was all 0's, it wouldn't be possible to do this. Similarly, if an ICMP packet wasn't reassembled properly, not only would the checksum likely be wrong, but you would most likely be able to tell exactly where the reassembly failed due to a gap or other inconsistency in the payload. This is just one example of why a payload with a sequence of bytes instead of all 0's is more useful to the user.

COM Port Commands CRC XOR

I have a check printer that I want to connect in Delphi 7 by COM port and operate.
I have a command that I extracted with Serial Port Monitor:
STX "PIRI(781" FS NULL ETX "0B" wich is 02 50 49 52 49 28 37 38 31 1c 00 03 30 42 hex
The manual says the following:
CRC (which is the last two digits after the ETX) - packet checksum. It
is calculated by the following algorithm: executing XOR for every
byte of the block including ETX by excluding STX. The data of the
checksum take up two bytes and are a symbolic representation of the
numeric in a hexadecimal calculation system.
I tried to an ONLINE CRC calculator and return a 1B result and a 27 numeric.
How to do it? For "PIRI(781" FS NULL ETX it should be 0B
The documentation incorrectly identifies the check value as a CRC. It is not. It is simply the exclusive-or of the noted bytes. The exclusive-or of 50 49 52 49 28 37 38 31 1c 00 03 is 0b. You then convert the 0b to hex (with an upper case B, i.e. 0B), and get 30 42.

Decode hex string encoding

I have a .bin saved with a VB program, the .bin format is:
String bytes | String
06 00 | C0 E1 E0 E8 F1 E0
The problem is I don't know how the string is encoded. I know what the string is supposed to be: Abaira
Can anyone recognize the encoding used?
I'm not aware of any standard character encoding for this. It is neither ASCII nor EBCDIC.
It seems to be some trivial sort of 8-bit (non-Unicode) ASCII (perhaps ANSI) encryption. Compare your unknown encoding with ASCII:
Unknown ASCII
Hex MSB LSB Hex MSB LSB
A CO 1100 0000 41 0100 0001
b E1 1110 0001 62 0110 0010
a E0 1110 0000 61 0110 0001
i E8 1110 1000 69 0110 1001
r F1 1111 0001 72 0111 0010
a E0 1110 0000 61 0110 0001
Let's define:
MSB: First nibble = most significant 4 bits
LSB: Second nibble = least significant 4 bits
_U: of Unknown
_A: of ASCII
Then you find:
MSB_U = MSB_A Xor 0x80 (maybe MSB_A Or 0x80)
LSB_U = LSB_A + 1 (to tell how overflow is handled I need to see ASCII char 'O' or 'o')
Then U is the concatenation MSB_U & LSB_U.
Further example ASCII to Unknown:
ASCII Hex MSB LSB MSB Xor 0x80 LSB - 1 Concatenated Hex
H 48 0100 1000 1100 1001 1100 0111 C7
e 65 0110 1001 1110 1010 1110 1000 E8
r 72 0111 0010 1111 0011 1111 0001 F1 (as you have shown)
b 62 0110 0010 1110 0011 1110 0001 E1 (do.)

Print Bitmap to ESC/POS printer

I am trying to print to a ESC/POS compatible printer and am struggling to get my head around GS v 0. I have just connected from a Mac and sending commands is hex via CoolTerm.
The docs say ...
GS v 0 m xL xH yL yH d1....dk
-----------------------------------------------------
[Name] Print raster bit image
[Format] ASCII GS v 0 m xL xH yL yH d1....dk
Hex 1D 76 30 m xL xH yL yH d1....dk
Decimal 29 118 48 m xL xH yL yH d1....dk
[Range] 0≤xL≤48, xH=0; 0≤yL≤255, yH=0; 0≤d≤255
k=(xL+xH×256)×(yL+yH×256)(k≠0)
[Description] Selects Raster bit-image mode. The value of m selects the mode, as follows:
+------+------------+----------------------------+---------------------------+
| m | MODE | Vertical Dot Density | Horizontal Dot density |
+------+------------+----------------------------+---------------------------+
|0, 48 | Normal | 200 DPI | 200 DPI |
+------+------------+----------------------------+---------------------------+
|1, 49 |Double-width| 200 DPI | 100 DPI |
+------+-------------+---------------------------+---------------------------+
|2, 50 |Double-height| 100 DPI | 200 DPI |
+------+-------------+---------------------------+---------------------------+
|3, 51 | Quadruple | 100 DPI | 100 DPI |
+------+-------------+---------------------------+---------------------------+
• xL, xH, select the number of data bits ( xL+ xH × 256) in the horizontal direction for the bit image.
• yL, yH, select the number of data bits ( yL+ yH × 256) in the vertical direction for the bit image.
• This command has no effect in all print modes (character size, emphasized, double-strike, upside-down, underline, white/black reverse printing, etc.) for raster bit image.
• The part of bit image that exceeds the printable area will not be printed.
• d indicates the bit-image data. Set time a bit to 1 prints a dot and setting it to 0 does not print a dot.
So from this I deduce I need to send the following in HEX
1D 76 30 30 20 00 00 01
Does the image data now follow this, and do I have to send a message saying the image has ended?
I remember there is <ESC>K (not GS) command to print 8 lines of pixels. See ESC commands for details. After K must be sent 2 bytes - number of data bytes and the real data. But it is not guaranteed whether every printer will support this. What is the brand and model of your?
The docs say:
Hex 1D 76 30 m xL xH yL yH d1....dk
with m being 0x48..0x51, so you can't send
1D 76 30 30 20 00 00 01
but rather, for instance
1D 76 30 48 20 00 00 01
I don't think it makes much sense that xH and yL are 0.

Assembly Language: Memory Bytes and Offsets

I am confused as to how memory is stored when declaring variables in assembly language. I have this block of sample code:
val1 db 1,2
val2 dw 1,2
val3 db '12'
From my study guide, it says that the total number of bytes required in memory to store the data declared by these three data definitions is 8 bytes (in decimal). How do I go about calculating this?
It also says that the offset into the data segment of val3 is 6 bytes and the hex byte at offset 5 is 00. I'm lost as to how to calculate these bytes and offsets.
Also, reading val1 into memory will produce 0102 but reading val3 into memory produces 3132. Are apostrophes represented by the 3 or where does it come from? How would val2 be read into memory?
You have two bytes, 0x01 and 0x02. That's two bytes so far.
Then you have two words, 0x0001 and 0x0002. That's another four bytes, making six to date.
The you have two more bytes making up the characters of the string '12', which are 0x31 and 0x32 in ASCII (a). That's another two bytes bringing the grand total to eight.
In little-endian format (which is what you're looking at here based on the memory values your question states), they're stored as:
offset value
------ -----
0 0x01
1 0x02
2 0x01
3 0x00
4 0x02
5 0x00
6 0x31
7 0x32
(a) The character set you're using in this case is the ASCII one (you can follow that link for a table describing all the characters in that set).
The byte values 0x30 thru 0x39 are the digits 0 thru 9, just as the bytes 0x41 thru 0x5A represent the upper-case alpha characters. The pseudo-op:
db '12'
is saying to insert the bytes for the characters '1' and '2'.
Similarly:
db 'Pax is a really cool guy',0
would give you the hex-dump representation:
addr +0 +1 +2 +3 +4 +5 +6 +7 +8 +9 +A +B +C +D +E +F +0123456789ABCDEF
0000 50 61 78 20 69 73 20 61 20 72 65 61 6C 6C 79 20 Pax is a really
0010 63 6F 6F 6C 20 67 75 79 00 cool guy.
val1 is two consecutive bytes, 1 and 2. db means "direct byte". val2 is two consecutive words, i.e. 4 bytes, again 1 and 2. in memory they will be 1, 0, 2, 0, assuming you're on a big endian machine. val3 is a two bytes string. 31 and 32 in are 49 and 50 in hexadecimal notation, they are ASCII codes for the characters "1" and "2".

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