Please is this correct formation

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i have to send the following packet format to a machine:-

C++

byte     byte     bit         bit
                       2       2         1           1
                 +---------+-------+-------------+----------+
                 |  CM     |   00  |  check sum  |  CR      |
                 +---------+-------+-------------+----------+

I have the following instructions:-

All data isASCII character code without checksumand CR.
Calculate sum total from top till before checksum ( 1 bite unit ),then recognize this complement of 1 as checksum.
Finish of format is CR (Hexadecimal: OD)

So to implement this
i took
This is main .cpp

C++

char data[6]="CM00";
typedef unsigned short int word16;
word16      check1;
int main()
{
int h=2;
check1 =calc_check(hh);
	data[4]= check1;
	data[5]=0x0D;
}

this is the file checksum .cpp

C++

typedef unsigned char      byte;    // Byte is a char
typedef unsigned short int word16;  // 16-bit word is a short int
typedef unsigned int       word32;  // 32-bit word is an int

//----- Defines ---------------------------------------------------------------
#define BUFFER_LEN        6      // Length of buffer
extern char data[6];

word16      check;  
//----- Prototypes ------------------------------------------------------------
word16 checksum(char *addr, word32 count);
//int calc_check(int);
//===== Main program ==========================================================
int calc_check(int w)
{
	//byte        buff[BUFFER_LEN]; // Buffer of packet bytes

	// 16-bit checksum value
	word32      i;                // Loop counter

	// Load buffer with BUFFER_LEN random bytes
	for (i=0; i<BUFFER_LEN; i++)
	{
		//buff[i] = (byte) rand();
		data[i]=(byte) rand();
	}

	// Compute the 16-bit checksum
	//check = checksum(buff, BUFFER_LEN);
	check = checksum(data, BUFFER_LEN);

	// Output the checksum
	printf("checksum = %04X \n", check);
	return check;

}

//=============================================================================
//=  Compute Internet Checksum for count bytes beginning at location addr     =
//=============================================================================
word16 checksum(char *addr, word32 count)
{
	register word32 sum = 0;
	word16 * waddr = reinterpret_cast<word16*>(addr);//added
	// Main summing loop
	while(count > 1)
	{
		//sum = sum + *((word16 *) addr)++;//main
		sum = sum + *waddr++;
		//sum += *((word16 *) addr)++;
		count = count - 2;
	}

	// Add left-over byte, if any
	if (count > 0)
		sum = sum + *((byte *) addr);

	// Fold 32-bit sum to 16 bits
	while (sum>>16)
		sum = (sum & 0xFFFF) + (sum >> 16);

	return(~sum);
}

I think i am doing something wrong in checksum.cpp as i have given the statement
Calculate sum total from top till before checksum so acc to this statement i should do

C++

for (i=0; i<2; i++)
    {
        //buff[i] = (byte) rand();
        data[i]=(byte) rand();
    }

Then am i doing this correct

C++

data[4]= check1;
	data[5]=0x0D;

Posted 12-Sep-12 5:15am

Tarun Batra

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Tarun Batra 13-Sep-12 2:09am

sir as u can see in the above packet format the check sum field is one bit,and i have done data[4]= check1; so is it correct?

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OriginalGriff · Answer 1 · 2012-09-12T05:41:00

Solution 1

No.
When you calculate your checksum, you overwrite everything in the buyffer, and then you are including all the characters in the message:

#define BUFFER_LEN        6      // Length of buffer
extern char data[6];
int calc_check(int w)
{
...
	// Load buffer with BUFFER_LEN random bytes
	for (i=0; i<BUFFER_LEN; i++)
	{
		data[i]=(byte) rand();
	}

	// Compute the 16-bit checksum
	check = checksum(data, BUFFER_LEN);
... 
}

Since the loop will overwrite the data you loaded before you called it.
Then you try to include the checksum byte in itself, and the CR terminator!

And that's ignoring that your code won't compile, because "h" and "hh" are not the same...

Why not hand the buffer and a length to the check routine, and have it return the check byte? If you did that, you could use the same function for incoming messages as well as outgoing...

Posted 12-Sep-12 5:41am

OriginalGriff

Comments

Tarun Batra 13-Sep-12 2:05am

sir to avoid the loop to overwrite the data I loaded before I called it,i should do something like this:-for (i=0; i<4; i++)
{
//buff[i] = (byte) rand();
data[i]=(byte) rand();
} and then i should do data[4]= check1;
data[5]=0x0D; as i have the following statement "Calculate sum total from top till before check sum then recognize this
complement of 1 as check sum. Now am i correct?

OriginalGriff 13-Sep-12 3:32am

No. Why the heck are you filling it with random data at all? All that does is overwrite whatever messgae you had to start with, and generates a check on the random rubbish!

Tarun Batra 13-Sep-12 3:37am

yes air iyou are correct i should comment this part -for (i=0; i<4; i++)
{
//buff[i] = (byte) rand();
data[i]=(byte) rand();} and sir can u tell me one more thing?

OriginalGriff 13-Sep-12 3:45am

It's not a case of commenting it - it probably shouldn't be there at all!

And what is the "other thing"?

Tarun Batra 13-Sep-12 3:50am

Ya sir you are correct,just tell me please that in this statement "check = checksum(data, BUFFER_LEN);" in check variable i will be getting the checksum value and as u can see i have data part as "char data[6]="CM00";" so is it correct data[4]= check; and acc to statement Finish of format is CR (Hexadecimal: OD) i am doing "data[5]=0x0D;",,,are these both correct?

OriginalGriff 13-Sep-12 4:06am

Almost - I wouldn't hand the BUFFER_LEN to the checksum function though. I would hand the length of the data in the buffer to be included in the checksum rather than the total size of the buffer.

In addition I would not play around with converting the char pointer to a word pointer - it may give unpredicable effects since char* data can be aligned on boundaries which may be illegal for word* data. Stick with char* instead.

Tarun Batra 13-Sep-12 4:20am

sir i wanna show u my code,can i have your mail id or should i display here?And what about
"Finish of format is CR (Hexadecimal: OD)" would u say am i doing correct?

OriginalGriff 13-Sep-12 4:33am

"can i have your mail id" - No :laugh:
I get enough Emails every day already!

"should i display here"
Relevant fragments, yes.

"would u say am i doing correct?" Yes - but I would use a #define instead of a "magic number".
#define FORMAT_TERMINATOR ((byte) 0x0D)
or similar. That way if you are wrong, or it changes later, it is easier to update the whole program.

Tarun Batra 13-Sep-12 5:32am

char data[6]="CM00";check2=calc_check(hh);int calc_check(int w)
{word32 i; har buff[4];
for (i=0; i<4; i++)
{
buff[i]=data[i];
}

char check11=0;
check11 = checksum(buff, 4);return check11;

}word16 checksum(char *addr, word32 count)
{
register word32 sum = 0;
char sum1=0;
word16 * waddr = reinterpret_cast<word16*>(addr);while(count > 1)
{sum1 = sum1 + *waddr++;count = count - 2;
}if (count > 0)
sum1= sum1 + *((byte *) addr);while (sum1>>24)
sum1 = (sum1 & 0xFFFF) + (sum1 >> 24);
printf("size of sum1 is %d\n",sizeof(sum1));
return(~sum1);}data[4]= check2;data[5]=FORMAT_TERMINATOR Please check now and tell me the changes

OriginalGriff 13-Sep-12 5:47am

Do you have a major wish to complicate things? :laugh:
char checksum(char* addr, int count)
{
char sum = 0;
while (count-- > 0)
{
sum += *addr++;
}
// Not sure what this bit is all about.
}

Your checksum calculation code looks very odd - you accumulate a sum in a char (8 bits long) and then try to "fold" it as a 32 bit quantity (so using 24 bits which don't actually exist, and which will all be sign extended to 1 or 0).

Do you have a description of your checksum calculation, because your text in the question is a bit...well...rubbish?

Tarun Batra 13-Sep-12 5:50am

I was trying to calculate the acc to this http://tools.ietf.org/rfc/rfc1071.txt,as if u have any code for standard check sum please give as i have to try many check sum algo.

OriginalGriff 13-Sep-12 13:59pm

Are you sure that is the right checksum calculator?
It's kind of old, and it generates a 16 bit result which you need to fit into an 8 bit field...

Tarun Batra 14-Sep-12 1:55am

Can u tell how to convert it into 8bit field?

OriginalGriff 14-Sep-12 3:32am

You can't - the algorithm is for a 16 bit checksum, not a 8. If you "convert it to 8 bit" then it is not the correct checksum for teh algorithm.

I suspect that you want a different checksum method - and there are loads of them - probably either a simple sum, or an exclusive OR which are the simplest, and most common. Do you have a specification for the device you are trying to interface with? It should give you all the details, not rely on a 24 year old document that may or may not be relevant! :laugh: