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I was given a challenge for Perl;	by caffine-iv	2003-01-19 02:34:14
Interesting...	by Spazmatic	2003-01-19 09:40:41
Stavros' XOR solution doesn't risk overflows	by kahuana	2003-01-19 09:52:32
Overflow isn't really a problem.	by Beorn	2003-01-19 11:37:40
The addition-subtraction method works for all integers, because the numbers wrap around. Example with single-byte valiables treated as natural numbers: `a = 250;` `b = 10;` `a = a + b;` a is now (250 + 10) mod 256 = 4. `b = a - b;` b is now (4 - 10) mod 256 = 250. `a = a - b;` a is now (4 - 250) mod 256 = 10. I'm not sure how it would work with floating point variables. The xor method on the other hand works on any fixed-size type.
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