I need to create EAN 8 bar code programmatically. I search an algorithm to calculate the checksum digit.
+3
A:
The algorithm is covered in this wikipedia article on EAN, note that EAN-8 is calculated in the same way as EAN-13.
Here's a worked example from http://www.barcodeisland.com/ean8.phtml :
Assuming we wish to encode the 7-digit message "5512345", we would calculate the checksum in the following manner:
Barcode 5 5 1 2 3 4 5
Odd/Even Pos? O E O E O E O
Weighting 3 1 3 1 3 1 3
Calculation 5*3 5*1 1*3 2*1 3*3 4*1 5*3
Weighted Sum 15 5 3 2 9 4 15
The total is 15 + 5 + 3 + 2 + 9 + 4 + 15 = 53. 7 must be added to 53 to produce a number evenly divisible by 10, thus the checksum digit is 7 and the completed bar code value is "55123457".
string code="55123457";
int sum1 = code[1] + code[3] + code[5]
int sum2 = 3 * (code[0] + code[2] + code[4] + code[6]);
int checksum_value = sum1 + sum2;
int checksum_digit = 10 - (checksum_value % 10);
if (checksum_digit == 10)
checksum_digit = 0;
Paul Dixon
2009-07-16 10:18:37
+3
A:
int checkSum(const std::vector<int>& code) const
{
if (code.size() < 8) return false;
for( SIZE_T i = 0; i< code.size(); i++ )
{
if( code[i] < 0 ) return false;
}
int sum1 = code[1] + code[3] + code[5]
int sum2 = 3 * (code[0] + code[2] + code[4] + code[6]);
int checksum_value = sum1 + sum2;
int checksum_digit = 10 - (checksum_value % 10);
if (checksum_digit == 10) checksum_digit = 0;
return checksum_digit;
}
SuperPro
2009-07-16 10:26:48
this is incorrect - sum1 should be (code[1] + code[3] + code[5]), sum2 should be 3 * (code[0] + code[2] + code[4] + code[6])
Paul Dixon
2009-07-16 13:10:30
OK. Only tested with EAN13. Seems that start weights are different in EAN8 and EAN13...
SuperPro
2009-07-16 13:52:44
Digits are counted from the right, so because 8 is even and 13 is odd, they have different weights. Thanks for editing, downvote removed!
Paul Dixon
2009-07-16 14:06:37