Ruby bitwise xor 2 same length ascii string

Question

I am trying to do bitwise xor 2 ASCII strings of the same length in ruby. I have come to the following:

'A'.unpack('B*').zip('B'.unpack('B*')).map{|a, b| (a.to_i ^ b.to_i).to_s}.pack('b*')
 => "\003"

It works. However, when performing string of 2 or more chars, it behaves wierdly, i think because of the value is too big for Fixnum, see below:

'AA'.unpack('B*').zip('BB'.unpack('B*')).map{|a, b| (a.to_i ^ b.to_i).to_s}.pack('b*')
 => "_\003"

If I stop right after the xor, without the to_s part, I see that it is not doing the bitwise XOR correctly:

'AA'.unpack('B*').zip('BB'.unpack('B*')).map{|a, b| (a.to_i ^ b.to_i)}
 => [1515925259]

Can someone help? Or can someone suggest another way of doing this? Many thanks

You provided attempted code,, that's good. But also you should provide sample input and output. For which inputs, you are expecting what outputs and what is the logic of getting that transformation. — Arup Rakshit
– Arup Rakshit, Commented Mar 14, 2014 at 20:02

Matt · Accepted Answer · 2014-03-14 21:04:44Z

2

I think you want to unpack with C (8-bit unsigned) rather than B (bit string) since the xor operator ^ operates on numbers rather than strings:

'AA'.unpack('C*').zip('BB'.unpack('C*')).map { |a,b| a^b }.pack('C*')
# => "\x03\x03"

3 is what one would expect from xoring 65 ('A') with 66 ('B').

edited Mar 14, 2014 at 21:04

answered Mar 14, 2014 at 19:53

Matt

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