C uint16 integer or char formatted? [duplicate]

StackOverflow https://stackoverflow.com/questions/19381004

  •  30-06-2022
  •  | 
  •  

Question

I'm new on C programming and I'm testing some code where I receive and process an UDP packet formatted as follow:

UINT16 port1
UINT16 port2

The corresponding values on this test are:

6005
5555

If I print the whole packet buffer I get something like this:

u^W³^U><9e>^D

How can I break it and get the right values? I've tried uint16_t and unsigned short and even the function htons, but didn't work either I just get wrong numbers, I'm trying something like this:

int l = 0;
unsigned short *primaryPort = *(unsigned short*) &buffer[l]; 
AddToLog(logInfo, "PrimaryPort: %u\n", primaryPort);
l += sizeof(*primaryPort);
unsigned short *secondaryPort = *(unsigned short *) &buffer[l]; 
AddToLog(logInfo, "SecondaryPort: %u\n", secondaryPort);
l += sizeof(*secondaryPort);

What is the right or the best way to get these values? I think these two uint16 values are not integer formatted, I think it's just an uint16 char text on this characters u^W³^U><9e>^D, so perhaps I'm getting the wrong numbers because I'm processing like an integer. But I've tried everything and I just can't get the right values.

Was it helpful?

Solution

When I look at the buffer you receive, it indeed starts with 6005 (little Endian):

6005 is 0x1775 and the first two characters are u (0x75) and Ctrl-W (0x17). So you should get the int value by:

 unsigned short primaryPort = (buffer[l] & 0xff) | ((buffer[l+1] & 0xff) << 8 );

and the same for the secondary port.

Edit

It works because:

In an expression like buffer[l] & 0xff, buffer[i] is implicitly casted into an int (search for 'integer promotion' to learn more). Masking it with 0xff is neccessary if buffer is char *: Then buffer[i] is a signed char and thus, if the highest bit is set, it's value is negative (e.g. '\xff' is -1). Masking it makes sure that we get a positive value in the range 0 - 255 instead. Last but not least, the shift operator << 8 shifts the byte value to the next higher byte the calculated int. Finally the result of the complete operation is implicitly casted to unsigned short which can be done safely because the calculated int is less or equal to 0xffff.

Licensed under: CC-BY-SA with attribution
Not affiliated with StackOverflow
scroll top