xor: how come so slow?

Michele michele at nectarine.it
Wed Oct 15 13:19:01 CEST 2008


Hi,
I'm trying to encode a byte data. Let's not focus on the process of
encoding; in fact, I want to emphasize that the method
create_random_block takes 0.5s to be executed (even Java it's faster) on
a Dual-Core 3.0Ghz machine:

took 46.746999979s, avg: 0.46746999979s

Thus I suppose that the xor operation between bytes raise the execution
time to 0.5; why I suppose that?
Because in Python there's no support for bytes and even for xoring
bytes, so I used a workaround:
I cycle on the two strings to be xor-red
    for every char in the strings
        convert one char on integer and then xor them; (ord)
        insert one char in the result, transforming the previous integer
in char (chr)

I suppose that ord() and char() are the main problems of my
implementation, but I don't know either way to xor two bytes of data
(bytes are represented as strings).
For more information, see the code attached.

How should I decrease the execution time? 

Thank you


from __future__ import division
import random
import time
import sha
import os

class Encoder(object):
    def create_random_block(self, data, seed, blocksize):
        number_of_blocks = int(len(data)/blocksize)
        random.seed(seed)
        random_block = ['0'] * blocksize
        for index in range(number_of_blocks):
            if int(random.getrandbits(1)) == 1:
                block = data[blocksize*index:blocksize*index+blocksize]
                for bit in range(len(block)):
                    random_block[bit] =
chr(ord(random_block[bit])^ord(block[bit])) # workaround per fare xor
bit a bit di str; xor e' solo supportato per int -> ord
        return ''.join(random_block)

    
x = Encoder()
piece = os.urandom(1024*1024)
blocksize = 16384
t1 = time.time()
for l in range(100):
    seed = random.getrandbits(32)
    block = x.create_random_block(piece, seed, blocksize)
t2 = time.time()
print 'took ' + str(t2-t1) + 's, avg: ' + str((t2-t1)/100.0) + 's'



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