From: Kristof Bastiaensen <kristof@...>
Date: 2007-01-30T02:05:11+09:00
Subject: Re: "a string".xor("another string")

On Mon, 29 Jan 2007 15:08:08 +0000, Ken Bloom wrote:
> once again, the result wasn't any faster. If you want fast, your best bet
> is probably to write it in C.

Or use an existing C extension :)
Using NArray:

require 'narray'
require 'benchmark'

class String
   def xor1(other)
     if other.empty?
       self
     else
       a1        = self.unpack("c*")
       a2        = other.unpack("c*")

       a2 *= 2   while a2.length < a1.length

       a1.zip(a2).collect{|c1,c2| c1^c2}.pack("c*")
     end
   end

   def xor2(other)
      if other.empty?
         self
      else
         if other.length < self.length
            div, mod = self.length.divmod(other.length)
            other = other * div + other[0, mod]
         end

         a1 = NArray.to_na(self, "byte")
         a2 = NArray.to_na(other, "byte")

         (a1 ^ a2).to_s
     end
   end
 end

$a1 = "abcdefg" * 1000
$a2 = "hijkl" * 1000
Benchmark.bm do |x|
   x.report("xor1:") {$a1.xor1($a2)}
   x.report("xor2:") {$a1.xor2($a2)}
end
       user       system      total      real
xor1:  0.030000   0.000000   0.030000 (  0.029777)
xor2:  0.000000   0.000000   0.000000 (  0.000427)