building bytes in python

Question

input:

    print(struct.pack("b", 0)
    print(struct.pack("b", 0))
    print(struct.pack("I", 3))
    print(struct.pack("I", 0))
    print(struct.pack("I", 0))
    print((struct.pack('bbIII', 0, 0, 3, 0, 0)))

output:

   b'\x00'
   b'\x00'
   b'\x03\x00\x00\x00'
   b'\x00\x00\x00\x00'
   b'\x00\x00\x00\x00'
   b'\x00\x00\x00\x00\x03\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00'

Singed chars at the last row are returning 2bytes for some reason.I don't understand why the last bytearray is returning the longer value than what I expect, I would like to know a reason and a solution of it. Thanks.

duckboycool · Accepted Answer · 2021-04-28 02:25:29Z

1

The struct is using the default @ character to represent native size and alignment. This is what's causing the sizes to be different. See the docs.

You can instead use = which will use native byte order but standard size.

print(struct.pack("=b", 0))
print(struct.pack("=bbIII", 0, 0, 3, 0, 0))

b'\x00'
b'\x00\x00\x03\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00' #14 bytes instead of 16

answered Apr 28, 2021 at 2:25

duckboycool

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1 Comment

Mark Ransom Over a year ago

So in other words, the extra bytes are padding to meet some arbitrary alignment requirement. Nice catch.

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