I have to calculate and verify checksums for IP packets (I am writing a router in C). The struct that represents an IP header has a 16 bit checksum but the total number of bits in the struct is not evenly divisible by 16 -- it is 8 bits short.
My question is this. Do I read the struct 16 bits at a time and pad the last set of bits with zeroes to calculate the checksum?
If you had the wrong number of bits, you would pad zero bits to the end of the data.
However, IP calculates over the header fields and thus always has a multiple of 16 bits on which to calculate the checksum. TCP sometimes needs an extra byte of zero at the end of the data.
http://en.wikipedia.org/wiki/IPv4_header_checksum
Related
In the TCP Header, why is the Urgent pointer field 16 bits long when it points to a sequence number of 32 bits
Don't think of it as pointing to a sequence number. It's true that the sequence numbers are used to count the total number of bytes transferred, that they are 32 bits in length, and that they can wrap around; the Urgent pointer, however, is a pointer into the data.
The urgent pointer indicates the offset from the sequence number of the received TCP segment with the urgent bit set. Therefore, it's not necessary for it to represent a 32 bit number, but it also means that the urgent data must be within the 64k following that sequence number.
I was looking at answers in Simple serial point-to-point communication protocol and it doesn't help me enough with my issue. I am also trying to communicate data between a computer and an 8-bit microcontroller at first, then eventually I want to communicate the one microcontroller to about 40 others via wireless radio modules. Basically one is designated as a master and the rest are slaves.
speed is an issue
The issue at hand is speed. because communication of every packet needs to be done at least 4x a second back and forth between the master and each slave.
Let's assume baud rate for data is 9600bps. That's 960 bytes a second.
If I used 16-byte packets then: 40 (slaves) times 16 (bytes) times 2 (ways) = 640. Divide that into 960 and that would mean well more than 1/2 a second. Not good.
If I used 8-byte packets then: 40 (slaves) times 8 (bytes) times 2 (ways) = 320. Divide that into 960 and that would mean 1/3 second. It's so-so.
But the thing is I need to watch my baud because too high of baud might mean missed data at larger distances, but you can see the speed difference between an 8 and 16 byte packet.
packet format idea
In my design, I may have a need to transmit a number in the low millions so that will use 24-bits which fits in my idea. But here's my initial idea:
Byte 1: Recipient address 0-255
Byte 2: Sender address 0-255
Byte 3: Command
Byte 4-6: Data
Byte 7-8: 16-bit fletcher checksum of above data
I don't mind if the above format is adjusted, just as long as I have at least 6 bits to identify the sender and receiver (since I'll only deal with 40 units), and the data with command included should be at least 4 bytes total.
How should I modify my data packet idea so that even the device that just turned on in the middle of reception can be in sync with the next set of data? Is there a way without stripping a bit from each data byte?
Rely on the check sum! My packet would consists of:
Recipient's address (0..40) XORed with 0x55
Sender's address (0..40) XORed with 0xAA
Command Byte
Data Byte 0
Data Byte 1
Data Byte 2
CRC8 sum, as suggested by Vroomfondel
Every receiver should have a sliding window of the last seven received bytes. When a byte was shifted in, that window should checked if it is valid:
Are the two addresses in the valid range?
Is it a valid command?
Is the CRC correct?
Especially the last one should safely reject packets on which the receiver hopped on off-sync.
If you have less than 32 command codes, you may go down to six bytes per packet: 40[Senders] times 40[Receivers] times 32[Commands] evaluates to 51200, which would fit into 16 bits instead of 24.
Don't forget to turn off the parity bit!
Update 2017-12-09: Here a receiving function:
typedef uint8_t U8;
void ByteReceived(U8 Byte)
{
static U8 Buf[7]; //Bytes received so far
static U8 BufBC=0;
Buf[BufBC++] = Byte;
if (BufBC<7) return; //Msg incomplete
/*** Seven Byte Message received ***/
//Check Addresses
U8 Adr;
Adr = Buf[0] ^ 0x55; if (Adr >= 40) goto Fail;
Adr = Buf[1] ^ 0xAA; if (Adr >= 40) goto Fail;
if (Buf[2] > ???) goto Fail; //Check Cmd
if (CalcCRC8(Buf, 6) != Buf[6]) goto Fail;
Evaluate(...);
BufBC=0; //empty Buf[]
return;
Fail:
//Seven Byte Msg invalid -> chop off first byte, could use memmove()
Buf[0] = Buf[1];
Buf[1] = Buf[2];
Buf[2] = Buf[3];
Buf[3] = Buf[4];
Buf[4] = Buf[5];
Buf[5] = Buf[6];
BufBC = 6;
}
I am writing a computer program which utilizes input from some equipment which I seldom have availible in my office. In order to develop and test this program I am trying to use an Arduino board to simulate the communication from the actual equipment. To this effect I create datapackets on the Arduino and send them to my computeer over the serial port. The packets are formated as a header and a hexidecimal integer, representing some sensor data.
The header is supposed to contain a checksum (2's complement 256-modulo). I am however not sure how to calculate it. In the datasheet of the equipment (which communication I try to simulate), it is stated that I should first compute the sum all bytes in the packet, and then take the 256-modulo of the sum and perform a 8-bit two's complement on the result.
However, as I am a newbie to bits, bytes and serial communication, I do not understand the following:
1) Lets say that I want to send the value 5500 as two bytes (high byte and low byte). Then the high-byte is '15' and the low-byte is '7c' in hexidecimal encoding, which corresponds to 21 and 124 in decimal values. Do I then add the contributions 21 and 124 to the checksum before taking the 256-modulo, or do I have to do some bit-magic beforehand?
2) How do I perform a two's compliment?
Here is a code which should illustrate how I think. The idea is to send a packet with a header containing a byte which states the length of the packet, a byte which states the type of the packet, and a byte for the checksum. Then a two-byte integer value representing some sensor value is devided into a high-byte and a low-byte, and transmitted low-byte first.
int intVal;
byte Len = 5;
byte checksum;
byte Type = 2;
byte intValHi;
byte intValLo;
void setup(){
Serial.begin(9600);
}
void loop(){
intVal = 5500; //assume that this is a sensor value
intValHi = highByte(intVal);
intValLo = lowByte(intVal);
//how to calculate the checksum? I unsuccessfully tried the following
checksum = 0;
checksum = (Len+checksum+Type+intValHi+intValLo) % 256;
//send header
Serial.write(Len);
Serial.write(checksum);
Serial.write(Type);
//send sensor data
Serial.write(intValLo);
Serial.write(intValHi)
}
Thanks!
The first thing you should understand is that mod 256 is the same thing as looking at the bottom log(256) => 8 bits.
To understand this you have to first realize what the 'mod' operation does and how digits are represented in hardware.
Mod is the remainder after an old-school division (ie only with whole numbers).
eg 5%2 = 1
Digits in hardware are stored in 'bits' which can be interpreted as base 2 mathematics.
Thus if you want to take the mod operation of a power of 2 you don't actually have to do any math.
This is just like if you want to have the remainder of the power of 10, you just take the lower digits.
ie. 157 % 100 = 57.
This can be sped up by using the fact that bytes should overflow by themselves. This means that all you have to do to take %256 of a bunch of numbers is to add them to a single byte and the arduino will take care of the rest.
For twos compliment see this question:
What is “2's Complement”?
Can some one explain the math behind this.
2 RAMS
16-bits wide
16-bit words
32k 16 bit words is the maximum fill up of each.
How do I get the '32k 16-bit words' from the information provided?
While your question was nearly impossible to understand, I'll try to answer anyway:
32768 16-bit words makes 65536 bytes (8 bits per byte)
2^16 = 65536 = The number of bytes that can be addressed using a 16 bit address.
This is more of a theoretical question than an actual problem I have.
If I understand correctly, the sequence number in the TCP header of a packet is the index of the first byte in the packet in the whole stream, correct? If that is the case, since the sequence number is an unsigned 32-bit integer, then what happens after more than FFFFFFFF = 4294967295 bytes are transferred? Will the sequence number wrap around, or will the sender send a SYN packet to restart at 0?
The sequence number loops back to 0. Source:
TCP sequence numbers and receive
windows behave very much like a clock.
The receive window shifts each time
the receiver receives and acknowledges
a new segment of data. Once it runs
out of sequence numbers, the sequence
number loops back to 0.
Also see chapter 4 of RFC 1323.
It wraps. RFC 793:
It is essential to remember that the actual sequence number space is finite, though very large. This space ranges from 0 to 2**32 - 1. Since the space is finite, all arithmetic dealing with sequence numbers must be performed modulo 2**32. This unsigned arithmetic preserves the relationship of sequence numbers as they cycle from 2**32 - 1 to 0 again. There are some subtleties to computer modulo arithmetic, so great care should be taken in programming the comparison of such values. The symbol "=<" means "less than or equal" (modulo 2**32).
Read more: http://www.faqs.org/rfcs/rfc793.html#ixzz0lcD37K7J
The sequence number is not actually the "index of the first byte in the packet in the whole stream" since sequence numbers deliberately start at a random value (this is to stop a form of attack known as the TCP Sequence Prediction Attack).
No SYN is required, the sequence number simply loops back to zero again once it gets to the limit.