Unpack from struct python

Question

Hello there!
I want to trun some bytes into integer form.
I use the code but if i remove the '!' before 'hhl' then the result is not true. What is the reason?
'a' should be 1 2 3
from struct import *
a =unpack('! hhl', b'\x00\x01\x00\x02\x00\x00\x00\x03')
print(a)

Answer 1

Exclamation mark determines the byte order. In your case it is used to represent data in network byte order, because the byte sequence was packed that way . In other words, if you'd have another byte order, you could use format string without '!':
>>> a = unpack('hhl', b'\x01\x00\x02\x00\x03\x00\x00\x00')
>>> a
(1, 2, 3)
So, the byte sequence should have been packed using 'hhl' fomat string to keep the "traditional" order:
>>> b = pack('hhl',1,2,3)
>>> b
b'\x01\x00\x02\x00\x03\x00\x00\x00'

Answer 2

So if i want communicate with network it's sth like htons()? It checks my host byte order and since network byte order is big endian
, if my host be little endian then it fixes the order from big endian to little endian otherwise it's a no op. Am i right? Or maybe it does not check my host byte order. It only changes the byte order. I must first realize if my host byte order and network byte order are different then use it. Which one is true?