Description of the binary field is:

Caller number, expressed with compressed BCD code, and the surplus bits are filled with “0xF”

I have tried to print with struct format `'16c'`

and I get: `('3', 'x00', 'x02', 'x05', 'x15', 'x13', 'G', 'O', 'xff', 'xff', 'xff', 'xff', 'xff', 'xff', 'xff', 'xff')`

and if I use `'16b'`

i get `(51, 0, 2, 5, 21, 19, 71, 79, -1, -1, -1, -1, -1, -1, -1, -1)`

. And it is not correct, I should get phone number, and numbers above are invalid.

print struct.unpack_from('>16b', str(data.read()),offset=46)

Above is code that didn’t work and I get invalid numbers. With what format should I unpack that 16 byte field and how to convert BCD code ?

## Answer

BCD codes work with 4 bits per number, and normally encode only the digits 0 – 9. So each byte in your sequence contains 2 numbers, 1 per 4 bits of information.

The following method uses a generator to produce those digits; I am assuming that a 0xF value means there are no more digits to follow:

def bcdDigits(chars): for char in chars: char = ord(char) for val in (char >> 4, char & 0xF): if val == 0xF: return yield val

Here I use a right-shift operator to move the left-most 4 bits to the right, and a bitwise AND to select just the right-most 4 bits.

Demonstration:

>>> characters = ('3', 'x00', 'x02', 'x05', 'x15', 'x13', 'G', 'O', 'xff', 'xff', 'xff', 'xff', 'xff', 'xff', 'xff', 'xff') >>> list(bcdDigits(characters)) [3, 3, 0, 0, 0, 2, 0, 5, 1, 5, 1, 3, 4, 7, 4]

The method works with the `c`

output; you can skip the `ord`

call in the method if you pass integers directly (but use the `B`

unsigned variant instead). Alternatively, you could just read those 16 bytes straight from your file and apply this function to those bytes directly without using struct.