I'm writing a Python library to quickly parse RAW files from a Sony camera. To detect the file's byte order, I originally used this code:nn```pythonndef check_for_endian(data: bytes) -> tuple[int, int] | None:n if data[:2] == 0x4949: # "II" little-endiann return 1, 0n if data[:2] == 0x4D4D: # "MM" big-endiann return 2, 0n return Nonen```nnThat no longer appears to work in Python 3.13.9, so I changed it to unpack the first bytes with `struct`. Did comparing a bytes slice with a hexadecimal integer ever work, or was the original comparison invalid from the beginning?
2 Answers
A `bytes` slice and an integer are different types. In Python 3, `data[:2] == 0x4949` evaluates to `False`; it isn’t a valid way to compare the two-byte sequence with an integer. The simplest solution is to compare against a bytes literal:nn```pythonndef check_for_endian(data: bytes) -> tuple[int, int] | None:n if data[:2] == b"II":n return 1, 0n if data[:2] == b"MM":n return 2, 0n return Nonen```nnYou can also use `b"\x49\x49"` and `b"\x4d\x4d"` if you prefer hexadecimal notation. Your `struct` approach can work too, but unpacking is unnecessary when you only need to inspect the raw header bytes.
If you keep using `struct`, make sure you unpack only the bytes you need. For example, `struct.unpack_from("<H", data, 0)[0]` reads an unsigned two-byte value. Using `unpack_from` without limiting the format is not necessary here. Also, comparing with `b"II"` or `b"MM"` avoids questions about integer byte order entirely.
Thanks! I’ll switch to comparing against byte literals. I’m certain the old version seemed to work before, but I can’t reproduce that behavior now; the only recent system change was an operating-system upgrade.