Slightly disappointed at the "two missing values" solution:
First: one needs to realize that you can solve the "missing number" problem just as well with sums. So, if you're trying to find the "one missing number" between 1 and n, you simply subtract all values from n*(n+1)/2 (the sum of all said numbers) and you end up with the missing one. (using wrap-around semantics, you don't even need to have more bits of memory than you need for the xor solution!)
Second: A common way to solve a problem with two variables is to build a system of two equations. One can compute, in a single pass, what is "a+b" and "a^b" ; solve the system, and you get both variables with no additional lookups. Now, it's true that if you get a carry from the addition, this problem might not be solvable...but, if you can afford one extra bit for the potential carry, you're all set.
Sums increase in size, however. The nice part of the xor trick is your xor result will be the size of the base of the numbers (64 bits or whatever). If you are reading several tb of 64 bit integers from disk that could get larger than your memory (though unlikely, probably why you see this in kernels where memory might be more of a concern especially with embedded)
Ah, you're right, I didn't think it through. It does feel like there must be a way to play with the bits so that you can retrieve it in a single pass, though - that'd be a much more interesting problem :).
(of course, you can do it with sum + product, but that's going to be fairly expensive for large numbers)
First: one needs to realize that you can solve the "missing number" problem just as well with sums. So, if you're trying to find the "one missing number" between 1 and n, you simply subtract all values from n*(n+1)/2 (the sum of all said numbers) and you end up with the missing one. (using wrap-around semantics, you don't even need to have more bits of memory than you need for the xor solution!)
Second: A common way to solve a problem with two variables is to build a system of two equations. One can compute, in a single pass, what is "a+b" and "a^b" ; solve the system, and you get both variables with no additional lookups. Now, it's true that if you get a carry from the addition, this problem might not be solvable...but, if you can afford one extra bit for the potential carry, you're all set.