Binary, Bits and Hexidecimal

Question

I am developing code to decode handphone PDU.
I need to decode the hexidecimal in the PDU into binary bits first.
Then I need to take every 7 bits and form them into a character.
E.g. A201 will give 1010 0010 0000 0001
so the code will have to return 0100010 as ASCII for the first character, 1100000 as ASCII for the second character.
How can I go about doing that in the most efficient method.
I was thinking about converting to string first (0's and 1's), then manually extract out every 7 bits and do the processing. But that looks quite tedious.
Can anybody comment on this. Thanks!

Answer 1

Can't see where the 1100000 comes from.
Have you really got characters on 7 bit boundaries? That would be very unusual and inconvenient.
This kind of bit manipulation is best tackled with shift and mask operations of which this will be a particularly tiresome example. You need to load your hex data into an array of, I would suggest, ints, loading two hex digits at a time. Work in blocks of 7 bytes (7 bytes becomes 8 characters).
int[] codes = new int[7];
String hex = "a7882312228823"; // say
for(int i = 0; i < 7; i++)
     codes[i] = Integer.parseInt(hex.substring(2 * i, 2 * i + 2);Then you need to extract the characters by combining the appropriate bits from each byte of data.
char output = new  char[8[;
output[0] = (codes[0] >> 1);   // first 7 bits of code 0
output[1] = (codes[0] << 7) | (codes[1] >> 2); // bottom bit of first and 6 bits from second
output[2] = (codes[1] << 6) | (codes[2] >> 3); // bottom 2 bits of second and top 5 from third
... etc ...Of course it's a lot messier than this because your data might not be a multiple of 7 bytes and the data supply is obviously more complicated.