Question

Is there an easy way to read these integers in? I’d prefer a built in method, but I assume it is possible to do with some bit operations.
Cheers

edit
I thought of another way to do it that is different to the ways below and in my opinion is clearer. It pads with zeros at the other end, then shifts the result. No if required because shifting fills with whatever the msb is initially.

struct.unpack('<i','\0'+ bytes)[0] >> 8

Answer 1

Python’s struct module lets you interpret bytes as different kinds of data structure, with control over endianness.

If you read a single three-byte number from the file, you can convert it thus:

struct.unpack('<I', bytes + '\0')

The module doesn’t appear to support 24-bit words, hence the '\0'-padding.

EDIT: Signed numbers are trickier. You can copy the high-bit, and set the high bit to zero because it moves to the highest place of 4 bytes (the last \xff has it).:

struct.unpack('<i', bytes + ('\0' if bytes[2] < '\x80' else '\xff'))

Or, for python3 (bytes is a reserved word, checking a byte of a byte array gives an int):

struct.unpack('<i', chunk + ('\0' if chunk[2] < 128 else '\xff'))