C++ Encoding Numbers - c++

Question

I am currently working on sending data to a receiving party based on mod96 encoding scheme. Following is the request structure to be sent from my side:
Field Size Type
1. Message Type 2 "TT"
2. Firm 2 Mod-96
3. Identifier Id 1 Alpha String
4. Start Sequence 3 Mod-96
5. End Sequence 3 Mod-96
My doubt is that the sequence number can be greater than 3 bytes. Suppose I have to send numbers 123 and 123456 as start and end sequence numbers, how to encode it in mod 96 format . Have dropped the query to the receiving party, but they are yet to answer it. Can somebody please throw some light on how to go about encoding the numbers in mod 96 format.

Answer 1

Provided there's a lot of missing detail on what you really need, here's how works Mod-96 econding:
You just use printable characters as if they were digits of a number:
when you encode in base 10 you know that 123 is 10^2*1 + 10^1*2 + 10^0*3
(oh and note that you arbitrary choose that 1's value is really one: value('1') = 1
when you encode in base 96 you know that 123 is
96^2*value('1')+ 96^2*value('2')+96^0*value('3')
since '1' is the ASCII character #49 then value('1') = 49-32 = 17
Encoding 3 printable characters into a number
unsigned int encode(char a, char b, char c){
return (a-32)*96*96 + (b-32)*96 + (c-32);
}
Encoding 2 printable characters into a number
unsigned int encode(char a, char b){
return (b-32)*96 + (c-32);
}
Decoding a number into 2 printable characters
void decode( char* a, char*b, unsigned int k){
* b = k % 96 +32;
* a = k / 96 +32;
}
Decoding a number into 3 printable characters
void decode( char* a, char*b, char*c, unsigned int k){
* c = k % 96 +32;
k/= 96;
* b = k % 96 +32;
* a = k/96 +32;
}
You also of course need to check that characters are printable (between 32 and 127 included) and that numbers you are going to decode are less than 9216 (for 2 characters encoded) and 884736(for 3 characters encoded).
You know the final size would be 6 bytes:
Size 2 => max of 9215 => need 14 bits storage (values up to 16383 unused)
Size 3 => max of 884735 => need 17 bits storage (values up to 131071 unused)
Your packet needs 14+17+17 bits of memory (wich is 48 => exactly 6 bytes) bits storage just for Mod-96 stuff.
Observation:
Instead of 3 fields of sizes(2+3+3) we could have used one field of size(8) => we would finally use 47 bits ( but is still rounded up to 6 bytes)
If you still store each encoded number into a integer number of bytes you would use the same amount of memory (14 bits fits into 2 bytes, 17 bits fits into 3 bytes) used by storing chars directly.