Proper way to pack a string in python

Question

What is the correct way to pack a five-byte asci string into python so that it is 8-bytes and little endian? For example, something like:

from struct import pack
pack('<ccccc3x', 'David')

Or:

pack('<ccccc3x', b'D', b'a', b'v', b'i', b'd')
# this works but seems unreadable to me

What would be the correct (or simplest) way to do this? Ideally, something like:

>>> pack('<ccccc3x', *bytearray(b'David'))

But it seems I cannot pass a list or do variable-unpacking and I have to do a new positional argument for every character.

The best I can come up with is the following, but I'm hoping there's a better way to do it:

# why won't it allow me to use `c`(har) and only `b`(yte) ?
>>> pack('<5b3x', *bytearray(b'David'))
b'David\x00\x00\x00'

Update: seems like this might be the best way:

>>> pack('<5s3x', b'David')
b'David\x00\x00\x00'

Not sure though about all the different 'char'-ish types: s, b, and c from the struct page.

AKX · Accepted Answer · 2020-09-17 05:40:14Z

2

Endianness only is a thing for values more than a byte in size, so you don't need to worry about it here.

The most succinct way I can think of to pad a 5-byte bytestring to 8 bytes is

b'David'.ljust(8, b'\0')

answered Sep 17, 2020 at 5:40

AKX

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2 Comments

Mad Physicist Over a year ago

Really neat solution

David542 Over a year ago

@AKX that's a great idea! I actually was doing the pack because I had a large bytearray and I was just appending/extending values to it, but yea for a asci-string I think this is much better.

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