[SOLVED] Sharing variables in C.

stf92 · 04-19-2013, 05:27 AM

Hi:
I have variable chk which is type unsigned long. Let's suppose this variable occupies four bytes (if it does not that's another problem). The program transmits a file over a line byte by byte at the same time maintaining a checksum which it transmit when done transmitting the file. The problem is I must transmit chk one byte at a time. How do I get access to the four bytes making chk? If I could declare

unsigned long chk;
unsigned char byte0 (at the same location as chk);
unsigned char byte1
unsigned char byte2
unsigned char byte3

then byte0 would be the least significant byte of chk. Also, I could do

lsbyte= chk mod 256 with the same result but I do not want to do arithmetic. Can two variables be declared so that they share the same location?

Ramurd · 04-19-2013, 05:41 AM

There are two ways:

One is pointers: make 4 pointers, one for each byte.
byte0 = address of chk+(0 * sizeof char);
byte1 = address of chk+(1 * sizeof char);
byte2 = address of chk+(2 * sizeof char);
byte3 = address of chk+(3 * sizeof char);

I think that one is the most elegant;
On the other hand, you can also use the values by shifting around chk.

millgates · 04-19-2013, 05:49 AM

Quote:

Originally Posted by stf92

The problem is I must transmit chk one byte at a time.

Why?

Quote:

Originally Posted by stf92

How do I get access to the four bytes making chk?

How about

Code:

unsigned long chk;
unsigned char* bytes = &chk;
printf("%d\n", bytes[0]);
...

Quote:

Originally Posted by stf92

then byte0 would be the least significant byte of chk.

Actually, that depends on the system endianness. When sending data over network, you have to allways define your transmission protocol so that the code works regardless of the byte order of both machines.

Quote:

Originally Posted by stf92

Also, I could do
lsbyte= chk mod 256 with the same result but I do not want to do arithmetic.

You have to do arithmetic, be it an int arithmetic or pointer arithmetic.

Quote:

Originally Posted by stf92

Can two variables be declared so that they share the same location?

Yes, that's what pointers are for. See above.

theNbomr · 04-19-2013, 11:48 AM

You can isolate bytes within a multibyte word with 'and's and 'shift's:

Code:

unsigned long word;  // assume 4-byte word, per your spec.
unsigned int  chksum;

   word = 0x12345678;

   chksum =    word & 0x000000ff;
   chksum += ( word & 0x0000ff00 ) >>  8;
   chksum += ( word & 0x00ff0000 ) >> 16;
   chksum += ( word & 0xff000000 ) >> 24;

--- rod.

stf92 · 04-19-2013, 12:44 PM

Thank you guys, I now see there are quite a few solutions. Still another one, I think, is unions. For instance:

Code:

union checksum {
unsigned long chk;
unsigned char bytes[4];
}

Here checksum.chk and checksum.bytes share the same memory location.

theNbomr · 04-19-2013, 01:19 PM

The use of unions as a form of 'conversion' memory space will only work in this case if the order of the bytes in unimportant (such as the way I've computed a simple sum checksum in the prior article). In a more 'CRC-like' scenario, where shifts and X-ORs are incorporated, the algorithm is sensitive to the byte ordering. The C language does not specify byte order of multibyte words, so the code would be unportable.

--- rod.

stf92 · 04-19-2013, 11:54 PM

Here is the program, in case anyone cares.

Sergei Steshenko · 04-20-2013, 10:59 AM

Quote:

Originally Posted by stf92

Hi:
I have variable chk which is type unsigned long. Let's suppose this variable occupies four bytes (if it does not that's another problem). The program transmits a file over a line byte by byte at the same time maintaining a checksum which it transmit when done transmitting the file. The problem is I must transmit chk one byte at a time. How do I get access to the four bytes making chk? If I could declare

unsigned long chk;
unsigned char byte0 (at the same location as chk);
unsigned char byte1
unsigned char byte2
unsigned char byte3

then byte0 would be the least significant byte of chk. Also, I could do

lsbyte= chk mod 256 with the same result but I do not want to do arithmetic. Can two variables be declared so that they share the same location?

Quote:

I have variable chk which is type unsigned long. Let's suppose this variable occupies four bytes (if it does not that's another problem)

- you have to read https://en.wikipedia.org/wiki/Inttypes.h#inttypes.h - nowadays fixed number of bits data types are readily available.

Quote:

The problem is I must transmit chk one byte at a time. How do I get access to the four bytes making chk?

- no problem:

Code:

chk & (((uint32)0xffu) << 0u)
chk & (((uint32)0xffu) << 1u)
chk & (((uint32)0xffu) << 2u)
chk & (((uint32)0xffu) << 3u)

is portable.