Ordering of format strings in Python's struct [duplicate]

Question

There seems to be some ordering issues when using format strings in Python.

For example, I am trying to unpack, in order: 3 doubles, 1 float, 2 doubles. This should be 5*8+4 == 44 bytes in size.

Consider the following code:

import struct

A = struct.calcsize("dddfdd")
B = struct.calcsize("dddf")+struct.calcsize("dd")
C = struct.calcsize("dddddf")

For me, in Python 3.6.9 on Linux, I get that A == 48, meanwhile B == C == 44 (as I would expect).

This means that something like the following would fail, when packed_data is 44 bytes.

unpacked_data = struct.unpack('dddfdd',packed_data)

I can, instead, resort to:

unpacked_data1 = struct.unpack('dddf',packed_data1)
unpacked_data2 = struct.unpack('dd',packed_data2)

What gives? Is there a better way to do the unpacking?

It's because of alignment, use the = prefix for the format string. See docs.python.org/3/library/… — mkrieger1
– mkrieger1, Commented Jul 23, 2020 at 11:59

Phil Leh · Accepted Answer · 2020-07-23 12:06:18Z

1

Try to specify a byte order with your format string:

# assumes a native byte order, size and alignment.
# This means for alignment purposes, that it will align bytes to multiples of 8 bytes
A = struct.calcsize("dddfdd")

# if you prepend '=' infront of the format string, the order assumes system byte order and no alignment
A = struct.calcsize("=dddfdd") 
# A == 44

To check your system byteorder you can also use:

import sys
print(sys.byteorder)

As reference look here

answered Jul 23, 2020 at 12:06

Phil Leh

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2 Comments

Valentin Aslanyan Over a year ago

I see, thanks. Is it padding it to 8 bytes because I have a 64 bit system?

Phil Leh Over a year ago

Yes, that is correct. For more information look at e.g. the Wikipedia entry.

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