Understanding struct.pack in python 2.7 and 3.5+

Question

I am attempting to understand; and resolve, why the following happens:

$ python
>>> import struct
>>> list(struct.pack('hh', *(50,50)))
['2', '\x00', '2', '\x00']
>>> exit()
$ python3
>>> import struct
>>> list(struct.pack('hh', *(50, 50)))
[50, 0, 50, 0]

I understand that hh stands for 2 shorts. I understand that struct.pack is converting the two integers (shorts) to a c style struct. But why does the output in 2.7 differ so much from 3.5?

Unfortunately I am stuck with python 2.7 for right now on this project and I need the output to be similar to one from python 3.5

In response to comment from Some Programmer Dude

$ python
>>> import struct
>>> a = list(struct.pack('hh', *(50, 50)))
>>> [int(_) for _ in a]
Traceback (most recent call last):
 File "<stdin>", line 1, in <module>
ValueError: invalid literal for int() with base 10: ''

• Some programmer dude: Loop over the list and create a list converting all characters to integers? List comprehension would work well for that.
• ForceBru: These outputs are absolutely the same. ord('2') == 50, ord('\x00') == 0, this is just a matter of representation.
• Sam Hammamy: @Someprogrammerdude see edit
• Antti Haapala: If you'd printed the return value without list() you'd have noticed that the return value is '2\x002\x00' vs b'2\x002\x00'; and then you'd have asked about the difference between list('abc') in python 2 and list(b'abc') in Python 3.
• Jean-François Fabre: @ForceBru they're not the same. One is a list of of str objects, the other is a list of int objects, obtained by iterating on a bytes object.

Answer 1

in python 2, struct.pack('hh', *(50,50)) returns a str object.

This has changed in python 3, where it returns a bytes object (difference between binary and string is a very important difference between both versions, even if bytes exists in python 2, it is the same as str).

To emulate this behaviour in python 2, you could get ASCII code of the characters by appling ord to each char of the result:

map(ord,struct.pack('hh', *(50,50)))

• Sam Hammamy: I wish to be like you one day Sir!