Name: [1836] Neil Dickson
Member: 108 months
Authored: 10 videos
Description: Hi, I'm a performance optimization enthusiast with an interest for many sorts of tough challenges. I've started Code Cortex, a group working to make software performance easier to improve. My sites are: ...

Performance Optimization Tutorial: Episode 2: Bits and Bytes, Part 1 [ID:1623] (3/3)

in series: Performance Optimization Tutorial

video tutorial by Neil Dickson, added 09/09

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This video begins looking at available bit operations and what they do.

NOT, OR, AND, XOR, and TEST are examined, allowing us to set, reset, complement, and check individual bits, among a few other things. Next episode will examine operations related to bit-shifting and have our first experiment of the tutorial.

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4. anonymous Fri, 09 Oct 2009 10:56

Very cool! I liked the puns and the information was great :D Keep 'em comin'!


Thanks. The confusion was with the total number of inputs being used (with whatever value), but it was rather the the number of values(set, or cleared) being odd or even to determine the result.

Sorry about the confusion; the sentence got pretty long so the meaning might've gotten lost. Some examples to clarify and elaborate:

0 xor 1 has an odd number (i.e. 1) of inputs that are ones, so the output is 1.

If you have a 3-way XOR gate, the value of xor(0,1,1) is 0, since it has an even number (i.e. 2) of ones, but xor(0,0,1) is 1, since it has an odd number (i.e. 1) of ones.

If you had a 10-way XOR gate, xor(1,0,0,1,1,0,1,0,0,1) is 1, since it has an odd number (i.e. 5) of ones.

In a different context, suppose that you have an array of dwords where every value is duplicated an even number of times except one value, which is duplicated an odd number of times. If you XOR all of the values together, the final result will be the value that's duplicated an odd number of times; this follows from that XOR is commutative (a XOR b = b XOR a) and nilpotent (a XOR a = 0).

I think we'll mostly just be using XOR for stuff like zeroing registers and negating floating-point numbers, but hopefully this helps give a bit better insight into XOR. :)

I got confused with the definition of XOR, because:

odd number of inputs set(to one) results in 1, else 0

But having 0 xor 1 results in 1, where it is a even number inputs and also not all are set(to one), which forms part of the else part of the previous sentence (resulting in 0)

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