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by Thorham 1983 days ago
You're wrong. Remember that XOR works on individual bits. If what you're saying was true, then swapping bits that are both set to 1 would also fail, which means this algorithm wouldn't work at all.

Edit: Ignore this post, I misread the original post as saying swapping the same values would fail.
2 comments
alowe 1983 days ago
They're not saying swapping equal valued variables breaks it. It's when the pointer is the same, using the trick to swap a variable with itself will set the variable to 0.
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Thorham 1983 days ago
Yeah, I miss-read that. Thanks for pointing that out!
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dahart 1983 days ago
> swapping bits that are both set to 1 would also fail

No, swapping two 1 bits works fine. Work it out more slowly, the article covered this and why it always works.

(1,1) => (1^1,1)=(0,1) => (0,1^0)=(0,1) => (0^1,1)=(1,1)

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