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checksum

Hi,

I have to calculate the checksum using the function below. I am new to Labview and my c skills is a little rysty. Please help.

Thankyou.

 

24 bit  = 00001 + 0000000000000000000

Var1 (5bit msb)

Var2 (19bit)

 

Encoded little endian 32 bit from 0x1FFFC to 0x1FFFF

 

uint32_t checksum8(const void *buf, uint32_t length)

{

uint32_t n, checksum=0;

uint8_t *p = (uint8_t *)buf;

for (n=0; n <length; n++)

{

checksum += *p++;

if (checksum > 255) checksum -= 255;

}

return (~checksum&0xff);

 

Combined number = checksum * 16777216 + Var1 * 524288 + Var2

 

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Message 1 of 11
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Hi labvcj,

 

here's the inner loop to give you a starter:

check.png

Best regards,
GerdW


using LV2016/2019/2021 on Win10/11+cRIO, TestStand2016/2019
Message 2 of 11
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I'd use a quotient and remainder


"Should be" isn't "Is" -Jay
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Message 3 of 11
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Hi Jeff,

 

this is what the OP wants to implement:

if (checksum > 255) checksum -= 255;

I'd use a quotient and remainder

I thought so too at first, but what would be the result when checksum is equal to 255?

 

Best regards,
GerdW


using LV2016/2019/2021 on Win10/11+cRIO, TestStand2016/2019
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Hi Jeff,

 

some more explanation (too late to edit previous post):

- "(checksum > 255)" refers to a Q&R with 256 as divider…

- "checksum -= 255" refers to a Q&R with 255 as divider…

Best regards,
GerdW


using LV2016/2019/2021 on Win10/11+cRIO, TestStand2016/2019
Message 5 of 11
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Seems like a strange choice of algorithm for the checksum.

In a normal world, for an 8 bit checksum, you would get the byte array and use sum array elements on it (the result would be limited to 8 bits and would all be fine).

 

In this case, each time you overflow the 8 bits you effectively need to discard the overflowed bit add 1 to the checksum. 

It doesn't add any robustness to the checksum algo, and feels to me like a mistake.

 

How sure are we that the c++ code actually works?

 

0xDEAD

Message 6 of 11
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@GerdW wrote:

Hi Jeff,

 

some more explanation (too late to edit previous post):

- "(checksum > 255)" refers to a Q&R with 256 as divider…

- "checksum -= 255" refers to a Q&R with 255 as divider…


Sorry,  I should have been specific.  Tha line is equivalent to 

     checksum = 256 mod (checksum); 

 

Or, just use a U8 😄 and no math required.   

 

You know where to post your example.  

 

Better you than Christian.  😄


"Should be" isn't "Is" -Jay
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Message 7 of 11
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Hi Jeff,

 

I don't get your math…

 

How is

checksum = 256 mod (checksum);

equivalent to

if (checksum > 255) checksum -= 255;

???

 

Example:

checksum := 270 (before math routine).

Your result: checksum = (256 mod 270) = 256

OP wanted: checksum = 270-255 = 15

 

Best regards,
GerdW


using LV2016/2019/2021 on Win10/11+cRIO, TestStand2016/2019
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256 mod (270) is 14. Modulus is the remainder of the dividend ,270 / divisor, 256.

 

My C++ may be rusty but,  is that a bad thing for a  LabVIEW guy?


"Should be" isn't "Is" -Jay
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Message 9 of 11
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@GerdW wrote:

Hi Jeff,

 

I don't get your math…

 

How is

checksum = 256 mod (checksum);

equivalent to

if (checksum > 255) checksum -= 255;

???

 

Example:

checksum := 270 (before math routine).

Your result: checksum = (256 mod 270) = 256

OP wanted: checksum = 270-255 = 15

 


Jeff's nomenclature is backwards.  It really should be:

checksum mod 255

That would match what the OP stated.  If you just add bytes and ignore the overflow (the way much checksums work), it would be a mod 256.

 

 


GCentral
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