The first example of Section 5.7 of Websocket Protocol RFC 6455 reads:
A single-frame unmasked text message
0x81 0x05 0x48 0x65 0x6c 0x6c 0x6f (contains "Hello")
What are the first two octets in this example and why are they there?
Looking up the UTF-8 codes for the last five octets I can see how those bytes would translate to Hello.
The first two octets in this example contain metadata about the data frame and the payload it contains. The last five octets represent the payload or Hello.
The first two octets 0x81 0x05 are hexadecimal representations of two bytes or 16 bits of data. In binary they translate to: 1000 0001 0000 0101.
Using the Data Frame Diagram from the RFC you can see what these bits tell us about the data frame. To summarize:
1st bit (FIN): indicates this is the final fragment of the message being sent
5th-8th bits (opcode): the payload is a text message in UTF-8
9th-16th bits (payload length): the length of the payload is 5
Related
I want to create an encryption program for studies using the DES algorithm. If my data is more than 64 bytes I can use CBC or EBC mode to encrypt it. But what if my data file is less than 64 bytes? Should I add white spaces at the end of the file to fill it and get 64 bytes? What is a good approach?
The DES block size is 64 bits, or 8 bytes. If your message is less than 8 bytes, you need to pad it. A common scheme is to fill out the block by repeatedly appending the length of the padding. For example, if your message is 3 bytes, and you need to fill an 8-byte block, you need 5 bytes of padding. Append the value 5, 5 times, to the message:
// Message ABC + 5 bytes of padding
0x41 0x42 0x43 + 0x05 0x05 0x05 0x05 0x05
This implies you need at least one byte of padding, so if your message is a complete block, you need to add a whole extra block just for padding. To remove the padding, look at the value of the last byte, N, and verify that the last N bytes of the plain text have that same value.
I'm trying a few days to get the type of the CRC with the following hex stream (sniffed with wireshark):
The Hex data i sniffed:
0000001ec001075465737431323308557365726e616d650850617373776f7264d224
This should be the DATA in HEX:
0000001ec001075465737431323308557365726e616d650850617373776f7264
So the last 4 digits are the checksum, in this case d224
I used many code snippets (PHP, java), and some online checksum calcuation sites:
e.g.:
http://www.scadacore.com/field-applications/programming-calculators/online-checksum-calculator/
But I don't get the correct CRC value.
Thanks!
Update 1
Here are more hex streams with CRC included (the last 4 digits):
0000001dc001045465737409557365726e616d65310950617373776f726431cc96
0000001dc001045465737409557365726e616d65320950617373776f72643289d9
0000001dc001045465737409557365726e616d65330950617373776f726433b51c
0000001dc001045465737409557365726e616d65340950617373776f7264340347
0000001dc001045465737409557365726e616d65350950617373776f7264353f82
It appears to be the ARC CRC, polynomial 0x8005, reflected, zero initial value and no final xor, if I discard the initial 0000001d on each message, and take the CRC at the end to be put in the stream in little-endian order.
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Hello my first ever question on here, in need of bit of guidance.
I'm working on a packet sniffer mainly to decode SNMP PDUs, however I am not entirely sure where to go with it.
Simply put my packet sniffer can extract information from packets however I am interested in the data payload field. It is written in C++ and I am using winsock.
What way should I go about this? Are the SNMP fields encoded in basic encoding rules or will I have to delve into ASN.1?
I am only looking to decode those SNMP fields within the data payload field into human readable form. They are going to be dumped into a text file. So I will be looking at decoding OIDs also. I am verifying everything as I go along with Wireshark and using GETIF to query my SNMP node.
Any guidance is appreciated.
EDIT:
Thanks user1793963 very well explained. Sorry to all who have marked this as too broad.
To elaborate on my original question could anyone explain the initial part of the PDU itself.
Example: My program outputs these hex values 30 82 00 A3 02 01 00, which is SEQUENCE (30), LENGTH (82) and two other values. This is from a GetRequest PDU.
The GetResponse PDU shows these values 30 81 B7 02 01 00, SEQUENCE, 81 in LENGTH and another value.
Could someone explain the values marked in bold. If it uses the simple TLV structure what are the values representing? What I know is the start of the sequence (30) and the total PDU length (which is 82 and 81) and I know 02 01 00 are INTEGER 1 in LENGTH and VERSION 0 however I do not understand 00 A3 (GetRequest) and B7 (GetResponse). What do these values represent?
Many thanks.
I am also using Wireshark to check values, however they do not state the start of the PDU sequence
Update, 9 years later :-)
30 82 00 A3 and 30 81 B7 02 the 30 is data type = sequence. Following that are (long) length fields (> 127 bytes).
The rule for large numbers is that only the lower 7 bits in the byte are used for holding the value (0-127). The highest order bit is used as a flag
to let the recipient know that this number spans more than one byte. If more than two bytes are required to encode the number, all will have the top bit set apart from the last byte. Any number over 127 must be encoded using more than one byte.
SNMP packets are encoded in ASN.1, but it's a very simple protocol (at least for SNMP v1 and v2.c, I do not have a lot of experience with v3). It uses a simple TLV structure: type, length, value. For example the bytes 0x4 0x6 0x70 0x75 0x62 0x6c 0x63 are a string (type 4) with length 6 and value "public". You can find a list of types here.
I find it useful to write out packages like this:
1. 0x30 0x34
2. 0x2 0x1 0x1
3. 0x4 0x6 0x70 0x75 0x62 0x6c 0x63
4. 0xa2 0x27
5. 0x2 0x4 0x1 0x2 0x3 0x4
6. 0x2 0x1 0x0
7. 0x2 0x1 0x0
8. 0x30 0x19
9. 0x30 0x17
10. 0x6 0x8 0x2b 0x6 0x1 0x2 0x1 0x1 0x2 0x0
11. 0x6 0xb 0x2b 0x6 0x1 0x4 0x1 0x85 0x22 0xd5 0xf 0x97 0x54
This is a response on a get request where I requested the OID 1.3.6.1.2.1.1.2.0 (sysObjectID).
A list (type 0x30) with a length of 52 bytes
The version: SNMPv2.c (0=v1, 1=v2.c)
The community string: "public". (note how this is send in cleartext)
A SNMP get-response with length 39
The request ID, a 32 bit integer.
The error code, 0 means no error.
The error index.
A list with length 25
A list with length 23
An OID with length 8: 1.3.6.1.2.1.1.2.0
An OID with length 11: 1.3.6.1.4.1.674.10895.3028
As you can see integers and strings are easy, but OIDs are a bit trickier. First of all the first two parts ("1.3") are represented as a single byte (0x2b). They did this to make each message a few bytes shorter.
The second problem is representing numbers larger than 255. To do this SNMP uses only the 7 least significant bits to store data, the most significant bit is a flag to signal that the data continues in the next byte. Numbers lower than 128 are stored in a single byte.
0x7f
= 0 111 1111
= 127
0x85 0x22
= 1 000 0101, 0 010 0010
= 000 0101 010 0010
= 674
0xc0 0x80 0x80 0x80
= 1 100 0000, 1 000 0000, 1 000 0000, 0 000 0000
= 100 0000 000 0000 000 0000 000 0000
= 0x8000000
This method is also used if the length of a TLV field is larger than 127.
RFC1592 describes the structure of the messages, take a look at page 11 for a similar example.
I can also recommend using Wireshark to analyze packets, it does an excellent job of translating them to something readable.
There is possible to use very big frames with some ethernet cards. One case is 9k frames or jumbo frame and other case is super jumbo frame (as i know, up to 64k).
What is format of frame used for such huge packets?
As I know, for normal frames two formats used widely:
Preamble Start_byte dest_mac src_mac Ethertype/length Payload CRC Interframe gap
10101010x7 10101011 6 bytes 6 bytes 2 bytes 46–1500 bytes 4 bytes 12 bytes
In one case, the ethertype is used for length, and in second - for packet type. If this field is < 0x0600 (decimal 1536 bytes), this is a length; if >= 0x0600 - it is a type.
So, it looks impossible to store 9000 in this field as length. How length of jumbo and super jumbo frames is stored?
The format used for jumbos is the same. Despite this description, the Ethertype field is not normally used to store a length. Normally in the Ethertype field, you will not see a length; you will see a type. Types are specified by IANA here:
https://www.iana.org/assignments/ieee-802-numbers
Usually you'll see one of the following types from the table:
Ethertype Exp. Ethernet Description References
---------------- -------------- -------------------- ----------
2054 0806 - - ARP [IANA]
2048 0800 513 1001 Internet IP (IPv4) [IANA]
86DD IPv6 [IANA]
There are two reasons this works:
The hardware sending the packet doesn't depend on the Layer 2 length field to know the Layer 1 length.
Some Layer 3 packets such as ARP have a known size (at least, for a known combination of hardware/protocol address length, such as Ethernet/IP where it is normally used). IPv4/IPv6 packets have a length field in their own header.
I downloaded Hex Workshop, and I was told to read a .dbc file.
It should contain 28,315 if you read
offset 0x04 and 0x05
I am unsure how to do this? What does 0x04 mean?
0x04 is hex for 4 (the 0x is just a common prefix convention for base 16 representation of numbers - since many people think in decimal), and that would be the fourth byte (since they are saying offset, they probably count the first byte as byte 0, so offset 0x04 would be the 5th byte).
I guess they are saying that the 4th and 5th byte together would be 28315, but did they say if this is little-endian or big-endian?
28315 (decimal) is 0x6E9B in hexadecimal notation, probably in the file in order 0x9B 0x6E if it's little-endian.
Note: Little-endian and big-endian refer to the order bytes are written. Humans typical write decimal notation and hexadecimal in a big-endian way, so:
256 would be written as 0x0100 (digits on the left are the biggest scale)
But that takes two bytes and little-endian systems will write the low byte first: 0x00 0x01. Big-endian systems will write the high-byte first: 0x01 0x00.
Typically Intel systems are little-endian and other systems vary.
Think of a binary file as a linear array of bytes.
0x04 would be the 5th (in a 0 based array) element in the array, and 0x05 would be the 6th.
The two values in 0x04 and 0x05 can be OR'ed together to create the number 28,315.
Since the value you are reading is 16 bit, you need to bitshift one value over and then OR them together, ie if you were manipulating the file in c#, you would use something like this:
int value = (ByteArray[4] >> 8) | ByteArray[5]);
Hopefully this helps explain how hex addresses work.
It's the 4th and the 5th XX code your viewing...
1 2 3 4 5 6
01 AB 11 7B FF 5A
So, the 0x04 and 0x05 is "7B" and "FF".
Assuming what you're saying, in your case 7BFF should be equal to your desired value.
HTH
0x04 in hex is 4 in decimal. 0x10 in hex is 16 in decimal. calc.exe can convert between hex and decimal for you.
Offset 4 means 4 bytes from the start of the file. Offset 0 is the first byte in the file.
Look at bytes 4 and five they should have the values 0x6E 0x9B (or 0x9B 0x6E) depending on your endianess.
Start here. Once you learn how to read hexadecimal values, you'll be in much better shape to actually solve your problem.