What does bitwise XOR (exclusive OR) mean?

Question

I\'m trying to understand the binary operators in C# or in general, in particular ^ - exclusive or.

For example:

Given an array of positive intege

Answer 1

The definition of the XOR (exclusive OR) operator, over bits, is that:

0 XOR 0 = 0
0 XOR 1 = 1
1 XOR 0 = 1
1 XOR 1 = 0

One of the ways to imagine it, is to say that the "1" on the right side changes the bit from the left side, and 0 on the right side doesn't change the bit on the left side. However, XOR is commutative, so the same is true if the sides are reversed. As any number can be represented in binary form, any two numbers can be XOR-ed together.

To prove it being commutative, you can simply look at its definition, and see that for every combination of bits on either side, the result is the same if the sides are changed. To prove it being associative, you can simply run through all possible combinations of having 3 bits being XOR-ed to each other, and the result will stay the same no matter what the order is.

Now, as we proved the above, let's see what happens if we XOR the same number at itself. Since the operation works on individual bits, we can test it on just two numbers: 0 and 1.

0 XOR 0 = 0
1 XOR 1 = 0

So, if you XOR a number onto itself, you always get 0 (believe it or not, but that property of XOR has been used by compilers, when a 0 needs to be loaded into a CPU register. It's faster to perform a bit operation than to explicitly push 0 into a register. The compiler will just produce assembly code to XOR a register onto itself).

Now, if X XOR X is 0, and XOR is associative, and you need to find out what number hasn't repeated in a sequence of numbers where all other numbers have been repeated two (or any other odd number of times). If we had the repeating numbers together, they will XOR to 0. Anything that is XOR-ed with 0 will remain itself. So, out of XOR-ing such a sequence, you will end up being left with a number that doesn't repeat (or repeats an even number of times).