Question

Given a byte b of 8 bits, the following formula returns the bits of b “swapped” (0 and 7 are swapped, 1 and 6 are swapped, etc.):

(b * 0x0202020202 & 0x010884422010) % 1023

When I have a function implementing this hack, namely

def reverseBits(b):
    return (b * 0x0202020202 & 0x010884422010) % 1023

then I get an overflow:

OverflowError: cannot fit ‘long’ into an index-sized integer

How can I implement the bit-swapping hack in Python?

Answer 1

This seems to work as you expect:

def sb(b):
    return (b * 0x0202020202 & 0x010884422010) % 1023

def harness(i):
    print '{:4}: {} -> {}'.format(i,bin(i)[2:].zfill(8),bin(sb(i))[2:].zfill(8))    

for i in range(256):
    harness(i)

Prints:

   0: 00000000 -> 00000000
   1: 00000001 -> 10000000
   2: 00000010 -> 01000000
   3: 00000011 -> 11000000
   4: 00000100 -> 00100000
   5: 00000101 -> 10100000
   6: 00000110 -> 01100000
   ...
 248: 11111000 -> 00011111
 249: 11111001 -> 10011111
 250: 11111010 -> 01011111
 251: 11111011 -> 11011111
 252: 11111100 -> 00111111
 253: 11111101 -> 10111111
 254: 11111110 -> 01111111
 255: 11111111 -> 11111111

This algorithm (and others) are found here. As stated, this method only works for a BYTE, so you will need to use another method for a larger bit pattern.

Edit

BTW: You can do bit reversal without any fear of overflow or math problems and for larger bit fields by using string manipulation:

>>> w=32
>>> s=bin(1234567)[2:].zfill(w)
>>> rb=s[::-1]
>>> s
'00000000000100101101011010000111'
>>> rb
'11100001011010110100100000000000'

Then convert back to an int this way:

int(rb,2)