How to byte-swap a 32-bit integer in python?

Question

Take this example:

i = 0x12345678
print("{:08x}".format(i))
   # shows 12345678
i = swap32(i)
print("{:08x}".format(i))
   # should print 78563412

What would be the swap32-function()? Is there a way to byte-swap an int in python, ideally with built-in tools?

Answer 1

One method is to use the struct module:

def swap32(i):
    return struct.unpack("<I", struct.pack(">I", i))[0]

First you pack your integer into a binary format using one endianness, then you unpack it using the other (it doesn't even matter which combination you use, since all you want to do is swap endianness).

Answer 2

Big endian means the layout of a 32 bit int has the most significant byte first,

e.g. 0x12345678 has the memory layout

msb             lsb
+------------------+
| 12 | 34 | 56 | 78|
+------------------+

while on little endian, the memory layout is

lsb             msb
+------------------+
| 78 | 56 | 34 | 12|
+------------------+

So you can just convert between them with some bit masking and shifting:

def swap32(x):
    return (((x << 24) & 0xFF000000) |
            ((x <<  8) & 0x00FF0000) |
            ((x >>  8) & 0x0000FF00) |
            ((x >> 24) & 0x000000FF))

Answer 3

From python 3.2 you can define function swap32() as the following:

def swap32(x):
    return int.from_bytes(x.to_bytes(4, byteorder='little'), byteorder='big', signed=False)

It uses array of bytes to represent the value and reverses order of bytes by changing endianness during conversion back to integer.