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
         


        

Answer 1

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))