I'm trying to assemble proper track data given a CVC3 and a bunch of positional parameters. But the EMV C-2 Kernel book is about as obtuse as you could imagine (would it kill somebody to include an example!?!). Can anyone help work this example:
9f62 - pcvc3(t1) - Position of CVC3 in track1: 0x38 (4-6?)
9f63 - punatc(t1) - Unpredictable Number Track1 Pos: 0x3C6 (2-3 7-10?)
9f64 - natc(t1) - Digits in track1 ATC: 4
9f65 - pcvc3(t2) - Position of CVC3 in track2: 0x38 (4-6)
9f66 - punatc(t2) - Unpredictable Number Track2 Pos: 0x3C6 (2-3 7-10?)
9f67 - Digits in track2 ATC: 4
After successful checksum generation:
9f61 - track2 CVC3 - 2EF4
9f60 - track1 CVC3 - 609B
9f36 - ATC - 1E47
assuming the discretionary data field starts out as all 0s, how does it end up? The spec says this:
Convert the binary encoded CVC3 (Track2) to the BCD encoding of the corresponding number expressed in base 10. Copy the q least significant digits of the BCD encoded CVC3 (Track2) in the eligible positions of the 'Discretionary Data' in Track 2 Data. The eligible positions are indicated by the q non-zero bits in PCVC3(Track2).
I read that as: CVC3 = 0x609B = 24731 (so copy 731? What does BCD have to do with this? Or are they just saying "copy the 731 as bcd encoded to the byte array"?)
yes you are correct it is rather obtuse. you are correct that you would convert your p values (pCVC3, and PUNATC) to binary. (0011 1000, 0011 1100 0110 for your track1 p values) you then right align the proper values with the discretionary data. example
Bxxxxxxxxxxxxxxxx^ /^14111014010000000000
....000000000000000000CCC000
....00000000000000AAAA000UU0
so you say that you CVC3 for track 1 is 609B which is 24,731, since you PCVC3 is asking for only 3 characters youd set 731 in. your ATC is 1E47 which is 7,751. your PUNATC is asking for 4 digits so you'd use 7751. FYI... if the ATC is lower than the requested characters you'd pad with 0's. Your unpredictable number is even more tricky... so you make a 4 byte random number. convert it to a uint (base 10) and then mark the 8 most significant bytes as 0. example. lets say your random 4 bytes is 29A6 06AE. in base 10 that is 698,746,542. mark out the first 8 characters with 0. and you are left with 000,000,002. you'd place 02 in for you unpredictable number placment.. so.. all that said your track would look like this
Bxxxxxxxxxxxxxxxx^ /^14111014017751731020
the last character is equal to the length of the unpredicatable number (numeric) digits. which was 02.. so the last digit is 2 making your final track data
%Bxxxxxxxxxxxxxxxx^ /^1411014017751731022;
track2 is very similar. good luck with it. I understand your frustrations with this. :)