I'm trying to build an parser to deserialze into object. Socket will send byte into parser. For the length of field 22 POS Entry Mode will N3 and byte will be always 2 digit. How to get the value for this field ?
You read the ASCII value of this field, and convert it into an integer.
if it says N3 that means they are three digits numeric field, so if the value say 51, you cast it to 051 and send the ASCII equivalent
Field 22 is pos entry mode. It's 3 digit numeric value. If format is BCD then 2 bytes contains 4 digits[ 0 (padded) + 3 digit POS entry mode). If format is ascci then it is 3 byte.
Related
When I save data byte[] in SQL Server the value change and add 0x0 in the first of value
the correct value (0xFFD8FFE000104A46)
the incorrect value (0x0FFD8FFE000104A46494600010102004C)
0xF and 0x0F are the same number, both are hexadecimal notations of number 15 decimal. A byte contains two hexadecimal 'digits'. If the left most digit is 0, it doesn't affect the value, just like zero-hundred and fifteen is the same as fifteen. The notation with the leading 0 just prints all the bytes, the one without strips the leading zeros.
Where the 494600010102004C part is coming from I don't know.
I have input comprising five character upper-case English letters e.g ABCDE and I need to convert this into two character unique ASCII output.
e.g. ABCDE and ZZZZZ should both give two different outputs
I have converted from ABCDE into hex which gives me 4142434445, but from this can I get to a two character output value I require?
Example:
INPUT1 = ABCDE
Converted to hex = 4142434445
INPUT2 = 4142434445
OUTPUT = ?? Any 2 ASCII Characters
Other examples of INPUT1 =
BIRAL
BRMAL
KLAAX
So you're starting with a 5-digit base-26 number, and you want to squeeze that into some 2-digit scheme with base n?
All possible 1-5 digit base-26 numbers gives you a number space of 26^5 = 11,881,376.
So you want the minimum n where n^2 >= 11,881,376.
Which gives you 3446.
Now it's up to you to go and find a suitable glyph block somewhere in UTF where you can reliably block-out 3446 separate characters to act as your new base/alphabet. And construct a mapping from your 5-char base-26 ABCDE type number onto your 2-char base-3446 wierd-glyph number. Good luck with that.
There's not enough variety in ASCII to do this, since it's only 128 printable characters. Limiting yourself to 2-chars of ASCII means you can only address a number space of 16384.
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. :)
For Objective-C:
Hi everyone, I'm trying to convert a hex input into binary. For example, someone enters in :
A55
I want that to convert to
101001010101
I've tried looking through past posts and none seem to be working for me. Any help would be greatly appreciated.
Use a lookup table: there are only 16 possible characters in a HEX representation, each corresponding to a four-character binary code group. Go through the HEX character-by-character, obtain a lookup, and put it in the resultant NSString.
Here is a copy of the lookup table for you.
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
8 1000
9 1001
A 1010
B 1011
C 1100
D 1101
E 1110
F 1111
There are multiple options as to how to do lookups. The simplest way would be making a 128-element array, and placing NSStrings at the elements corresponding to codes of the characters (i.e. at positions '0', '1', ..., 'E', 'F', with single quotes; these are very important).
I believe there is a built-in function for this. If not, you should at least be able to go hex->dec then dec->bin
You can write the conversion from scratch if you know the number of characters, bin to hex is common enough algorithmically.
A mathematical look at the algorithms
SO Answers in C/C++ Another
Base 10 to base n in Objective C
C Hex->Bin
Build a lookup table (an array where you can supply a value between 0 and 15 to get the binary for that hex digit):
char *hex_to_bin[] = {
"0000", "0001", "0010", "0011",
/* ... */
"1100", "1101", "1110", "1111"
};
There should be 16 elements in that table. The conversion process for multiple digits is to handle one digit at a time, appending the results onto the end of your result storage.
Use getchar() to read a char:
int c = getchar();
if (c < 0) { puts("Error: Invalid input or premature closure."); }
Use strchr() to determine which array index to retrieve:
char *digits = "00112233445566778899AaBbCcDdEeFf";
size_t digit = (strchr(digits, c) - digits) / 2;
Look up the corresponding binary values for digit:
printf("%s", hex_to_bin[digit]); // You'll want to use strcat here.
I am faced with the task of sending ISO 8583 Rev 93 messages and am using openiso8583.net. The company that is consuming my messages gave message samples and I am unclear about the following Field attributes:
Special Characters
Alphabetic & Numeric Characters
Alphabetic & Special Characters
Number & Special Characters
Alphabetic, Numeric, & Special Characters
Here is the example:
Signon Reply
0810822000000200000004000000000000000501130427000005F0F00001
NUM |FLDNAME |FIELD DESCRIPTION |LEN |T|FIELD VALUE
-----|--------|-------------------------------|----|-|--------------------------
N/A |MSGTYPE |MESSAGE TYPE |F2 |H|0810`
N/A |BITMAP1 |FIRST BITMAP |B8 |H|8220000002000000`
1 |BITMAP2 |SECOND BITMAP |B8 |H|0400000000000000`
7 |MISDTMDT|TRANSMISSION DATE AND TIME |F5 |H|0501130427`
11 |MISDSTAN|SYSTEM TRACE AUDIT NUMBER |F3 |H|000005`
39 |MISDRSPC|RESPONSE CODE |F2 |C|00` <------?
70 |MISDNMIC|NETWORK MANAGEMENT INFO CODE |F2 |H|0001`
First, take a look at the message bytes:
0810822000000200000004000000000000000501130427000005*F0F0*0001
My question is how the two bytes { 0xF0, 0xF0 } translates to "00". If the company was sending ASCII, I would expect "00" to be { 0x30, 0x30 }. BCD is used for Numeric values but I can't seem to figure out how character values are being encoded.
Here is the description for field 39:
039:
Network Response Code
Attributes:
an 2*
Description:
A field that indicates the result of a previous related request. It will indicate
approval or reason for rejection if not approved. It is also used to indicate to the
device processor whether or not machines that are capable of retaining the customer's
card should do so.
Format:
In transaction replies, the response code must contain one of the following values
with their corresponding meanings. For debit/host-data-capture 0220 / 0420 messages, a
response code of '00' must be returned to indicate the transaction was approved. For
EBT transactions, please refer to section 4.8 EBT Transaction Receipt Requirements.
an2 means Alphabetic & Numeric Characters
Bitmap 1 is 64 bits
Bitmap 2 is 64 bits
Msg Type is 4 bytes
Field 7 is Numeric 4-bit BCD (Packed unsigned) 10, 5 bytes
Field 11 is Numeric 4-bit BCD (Packed unsigned) 6, 3 bytes
Field 39 is an 2, I assume 2 bytes
Field 70 is Numeric 4-bit BCD (Packed unsigned) 3, 2 bytes
Any clues or pointers would be greatly appreciated. Maybe someone knows of some encoding I clearly do not or can give a general explenation of how characters are encoded for ISO 8583 Rev 93. I do realize that each company can have different implementations though.
I hate answering my own questions quickly but...I just found the answer.
EBCDIC
I guess not being a programmer in the days of punch cards slowed me down on this one
0xF0 = '0'