Python, how to decode Binary coded decimal (BCD)

Question

Description of the binary field is:

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

Answer 1

Python 3.x has this built-in for bytes objects.

>>> c = b'3\x00\x02\x05\x15\x13GO\xff\xff\xff\xff\xff\xff\xff\xff'
>>> c.hex()
'330002051513474fffffffffffffffff'

And back:

>>> c = "9F1A020840"
>>> bytes.fromhex(c)
b'\x9f\x1a\x02\x08@'

Answer 2

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.