pythonbytebit

Python: Extracting bits from a byte


I'm reading a binary file in python and the documentation for the file format says:

Flag (in binary)Meaning

1 nnn nnnn Indicates that there is one data byte to follow that is to be duplicated nnn nnnn (127 maximum) times.

0 nnn nnnn Indicates that there are nnn nnnn bytes of image data to follow (127 bytes maximum) and that there are no duplications.

n 000 0000 End of line field. Indicates the end of a line record. The value of n may be either zero or one. Note that the end of line field is required and that it is reflected in the length of line record field mentioned above.

When reading the file I'm expecting the byte I'm at to return 1 nnn nnnn where the nnn nnnn part should be 50.

I've been able to do this using the following:

flag = byte >> 7
numbytes = int(bin(byte)[3:], 2)

But the numbytes calculation feels like a cheap workaround.

Can I do more bit math to accomplish the calculation of numbytes?

How would you approach this?


Solution

  • You can strip off the leading bit using a mask ANDed with a byte from file. That will leave you with the value of the remaining bits:

    mask =  0b01111111
    byte_from_file = 0b10101010
    value = mask & byte_from_file
    print bin(value)
    >> 0b101010
    print value
    >> 42
    

    I find the binary numbers easier to understand than hex when doing bit-masking.

    EDIT: Slightly more complete example for your use case:

    LEADING_BIT_MASK =  0b10000000
    VALUE_MASK = 0b01111111
    
    values = [0b10101010, 0b01010101, 0b0000000, 0b10000000]
    
    for v in values:
        value = v & VALUE_MASK
        has_leading_bit = v & LEADING_BIT_MASK
        if value == 0:
            print "EOL"
        elif has_leading_bit:
            print "leading one", value
        elif not has_leading_bit:
            print "leading zero", value