How are my bytes in C stored? - c

Question

First, I'm a student still. So I am not very experienced.
I'm working with a piece of bluetooth hardware and I am using its protocol to send it commands. The protocol requires packets to be sent with LSB first for each packet field.
I was getting error packets back to me indicating my CRC values were wrong so I did some investigating. I found the problem, but I became confused in the process.
Here is Some GDB output and other information elucidating my confusion.
I'm sending a packet that should look like this:
|Start Flag| Packet Num | Command | Payload | CRC | End Flag|
0xfc 0x1 0x0 0x8 0x0 0x5 0x59 0x42 0xfd
Here is some GDB output:
print /x reqId_ep
$1 = {start_flag = 0xfc, data = {packet_num = 0x1, command = {0x0, 0x8}, payload = {
0x0, 0x5}}, crc = 0x5942, end_flag = 0xfd}
reqId_ep is the variable name of the packet I'm sending. It looks all good there, but I am receiving the CRC error codes from it so something must be wrong.
Here I examine 9 bytes in hex starting from the address of my packet to send:
x/9bx 0x7fffffffdee0
0xfc 0x01 0x00 0x08 0x00 0x05 0x42 0x59 0xfd
And here the problem becomes apparent. The CRC is not LSB first. (0x42 0x59)
To fix my problem I removed the htons() that I set my CRC value equal with.
And here is the same output above without htons():
p/x reqId_ep
$1 = {start_flag = 0xfc, data = {packet_num = 0x1, command = {0x0, 0x8}, payload = {
0x0, 0x5}}, crc = 0x4259, end_flag = 0xfd}
Here the CRC value is not LSB.
But then:
x/9bx 0x7fffffffdee0
0xfc 0x01 0x00 0x08 0x00 0x05 0x59 0x42 0xfd
Here the CRC value is LSB first.
So apparently the storing of C is LSB first? Can someone please cast a light of knowledge upon me for this situation? Thank you kindly.

Answer 1

This has to do with Endianness in computing:
http://en.wikipedia.org/wiki/Endianness#Endianness_and_operating_systems_on_architectures
For example, the value 4660 (base-ten) is 0x1234 in hex. On a Big Endian system, it would be stored in memory as 1234 while on a Little Endian system it would be stored as 3412
If you want to avoid this sort of issue in the future, it might just be easiest to create a large array or struct of unsigned char, and store individual values in it.
eg:
|Start Flag| Packet Num | Command | Payload | CRC | End Flag|
0xfc 0x1 0x0 0x8 0x0 0x5 0x59 0x42 0xfd
typedef struct packet {
unsigned char startFlag;
unsigned char packetNum;
unsigned char commandMSB;
unsigned char commandLSB;
unsigned char payloadMSB;
unsigned char payloadLSB;
unsigned char crcMSB;
unsigned char crcLSB;
unsigned char endFlag;
} packet_t;
You could then create a function that you compile differently based on the type of system you are building for using preprocessor macros.
eg:
/* Uncomment the line below if you are using a little endian system;
/* otherwise, leave it commented
*/
//#define LITTLE_ENDIAN_SYSTEM
// Function protocol
void writeCommand(int cmd);
//Function definition
void writeCommand(int cmd, packet_t* pkt)
{
if(!pkt)
{
printf("Error, invalid pointer!");
return;
}
#if LITTLE_ENDIAN_SYSTEM
pkt->commandMSB = (cmd && 0xFF00) >> 8;
pkt->commandLSB = (cmd && 0x00FF);
# else // Big Endian system
pkt->commandMSB = (cmd && 0x00FF);
pkt->commandLSB = (cmd && 0xFF00) >> 8;
#endif
// Done
}
int main void()
{
packet_t myPacket = {0}; //Initialize so it is zeroed out
writeCommand(0x1234,&myPacket);
return 0;
}
One final note: avoid sending structs as a stream of data, send it's individual elements one-at-a-time instead! ie: don't assume that the struct is stored internally in this case like a giant array of unsigned characters. There are things that the compiler and system put in place like packing and allignment, and the struct could actually be larger than 9 x sizeof(unsigned char).
Good luck!

Answer 2

This is architecture dependent based on which processor you're targeting. There are what is known as "Big Endian" systems, which store the most significant byte of a word first, and "Little Endian" systems that store the least significant byte first. It looks like you're looking at a Little Endian system there.