How do you pack three bytes into a python struct?

Question

Given a hex value of 0x183C6 I'd like to write out its corresponding byte value using exactly 3 bytes. I know that struct would do this to 4 bytes but my constraints are to 3 bytes.

So

>>> val = 0x183c6
>>> struct.pack('>L', val)
'\x00\x01\x83\xc6'

Is great but I just want:

'\x01\x83\xc6'

I could post process and strip it after the fact, I could break it up and put each individual byte into a bytearray(), or perhaps do some bit shifting but I wonder if there is a easier way of doing it within struct without post processing.

Take a look at this http://stackoverflow.com/questions/34794640/python-struct-pack-pack-multiple-datas-in-a-list-or-a-tuple — Hans, Mar 28 '16 at 17:47

chepner · Accepted Answer · 2016-03-28T18:10:14.320

3

You can split the value into its high byte and low 16-bit word with divmod:

>>> struct.pack(">BH", *divmod(0x183c6, 1<<16))
'\x01\x83\xc6'

To unpack,

>>> struct.unpack(">I", '\x00' + '\x01\x83\xc6')

edited Mar 28 '16 at 18:10

answered Mar 28 '16 at 18:05

chepner

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But how would you unpack it? – Mark Nunberg Mar 28 '16 at 18:06
I have a byte array, so I'm doing: `struct.unpack(">I", b'\0' + data)`, for me `'\x00'` won't cut it. – Jari Turkia Jan 21 '21 at 20:18

score 1 · Answer 2 · answered Mar 28 '16 at 17:56

You can't really do that, since that's not how structs work. If you absolutely know that the value will never exceed 3 bytes, you can use big-endian encoding for both packing and unpacking:

>>> val = 0x183c6
>>> packed = struct.pack('>L', val)[1:]
>>> packed
'\x01\x83\xc6'
>>> '{0:x}'.format(struct.unpack('>L', '\x00' + packed)[0])
'183c6'

Note that the packed value will pretty much be useless for passing to other C functions, or similar.

How do you pack three bytes into a python struct?

2 Answers2