I would like to pack 0x12345678
into \x34\x12\x78\x56
I wrote this
>>> a = struct.unpack('cccc', struct.pack('I', 0x12345678))
>>> struct.pack('cccc', [a[i] for i in (1,0,3,2)])
But it is very ugly. Is there an easier way to do it?
Edit: Proper way: using short and reversing the endian type
import struct
a = struct.unpack('hh', struct.pack('I', 0x12345678))
struct.pack('>hh', *a)
Older answer
You can reverse the usage ...
import struct
a1, a0, a3, a2 = struct.unpack('cccc', struct.pack('I', 0x12345678))
struct.pack('cccc', a0, a1, a2, a3)
But it's making a lot of variables
Alternatively, swapping in the array will allow you to pass the result easier:
import struct
a = struct.unpack('cccc', struct.pack('I', 0x12345678))
b = sum([[a[i+1], a[i]] for i in range(0, len(a), 2)], [])
struct.pack('cccc', *b)
Note: Their is probably a better way for swapping :)